maths.free › Calculus › 5. Sequences and Series › The Divergence and Integral Tests

The Divergence and Integral Tests

Use the divergence test to demonstrate that a series diverges.

Divergence Test

A series \(\sum _{n=1}^{\infty }{a}_{n}\) being convergent is equivalent to the convergence of the sequence of partial sums (\({S}_{k}\)) as \(k\to \infty .\)

To verify this, notice that from the algebraic properties,

\[\underset{k\to \infty }{\text{lim}}{a}_{k}=\underset{k\to \infty }{\text{lim}}({S}_{k}-{S}_{k-1})=\underset{k\to \infty }{\text{lim}}{S}_{k}-\underset{k\to \infty }{\text{lim}}{S}_{k-1}=S-S=0.\]

Therefore, if \(\sum _{n=1}^{\infty }{a}_{n}\) converges, the \(n\text{th}\) term \({a}_{n}\to 0\) as \(n\to \infty .\) An important consequence of this fact is the following statement:

\[\text{If}\ {a}_{n}↛0\ \text{as}\ n\to \infty ,\sum _{n=1}^{\infty }{a}_{n}\ \text{diverges}.\]

This test is known as the divergence test because it provides a way of proving that a series diverges.

It is important to note that the converse of this theorem is not true. That is, if \(\underset{n\to \infty }{\text{lim}}{a}_{n}=0,\) we cannot make any conclusion about the convergence of \(\sum _{n=1}^{\infty }{a}_{n}.\) For example, \(\underset{n\to \infty }{\text{lim}}(1\text{/}n)=0,\) but the harmonic series \(\sum _{n=1}^{\infty }1\text{/}n\) diverges. In this section and the remaining sections of this chapter, we show many more examples of such series. Consequently, although we can use the divergence test to show that a series diverges, we cannot use it to prove that a series converges. Specifically, if \({a}_{n}\to 0,\) the divergence test is inconclusive.

Example

Try it.

For each of the following series, apply the divergence test. If the divergence test proves that the series diverges, state so. Otherwise, indicate that the divergence test is inconclusive.

\(\sum _{n=1}^{\infty }\frac{n}{3n-1}\)
\(\sum _{n=1}^{\infty }\frac{1}{{n}^{3}}\)
\(\sum _{n=1}^{\infty }{e}^{1\text{/}{n}^{2}}\)

Solution

Since \(n\text{/}(3n-1)\to 1\text{/}3\ne 0,\) by the divergence test, we can conclude that
\[\sum _{n=1}^{\infty }\frac{n}{3n-1}\]
diverges.
Since \(1\text{/}{n}^{3}\to 0,\) the divergence test is inconclusive.
Since \({e}^{1\text{/}{n}^{2}}\to 1\ne 0,\) by the divergence test, the series
\[\sum _{n=1}^{\infty }{e}^{1\text{/}{n}^{2}}\]
diverges.

Integral Test

In the previous section, we proved that the harmonic series diverges by looking at the sequence of partial sums \(\{{S}_{k}\}\) and showing that \({S}_{{2}^{k}}>1+k\text{/}2\) for all positive integers \(k.\) In this section we use a different technique to prove the divergence of the harmonic series. This technique is important because it is used to prove the divergence or convergence of many other series. This test, called the integral test, compares an infinite sum to an improper integral. It is important to note that this test can only be applied when we are considering a series whose terms are all positive.

To illustrate how the integral test works, use the harmonic series as an example. In , we depict the harmonic series by sketching a sequence of rectangles with areas \(1,1\text{/}2,1\text{/}3,1\text{/}4\text{,\ldots }\) along with the function \(f(x)=1\text{/}x.\) From the graph, we see that

\[\sum _{n=1}^{k}\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\text{\cdots }+\frac{1}{k}>{\int }_{1}^{k+1}\frac{1}{x}dx.\]

Therefore, for each \(k,\) the \(k\text{th}\) partial sum \({S}_{k}\) satisfies

\[{S}_{k}=\sum _{n=1}^{k}\frac{1}{n}>{\int }_{1}^{k+1}\frac{1}{x}dx=\text{ln}\ x\ {|}_{1}^{k+1}=\text{ln}(k+1)-\text{ln}(1)=\text{ln}(k+1).\]

Since \(\underset{k\to \infty }{\text{lim}}\text{ln}(k+1)=\infty ,\) we see that the sequence of partial sums \(\{{S}_{k}\}\) is unbounded. Therefore, \(\{{S}_{k}\}\) diverges, and, consequently, the series \(\sum _{n=1}^{\infty }\frac{1}{n}\) also diverges.

Now consider the series \(\sum _{n=1}^{\infty }1\text{/}{n}^{2}.\) We show how an integral can be used to prove that this series converges. In , we sketch a sequence of rectangles with areas \(1,1\text{/}{2}^{2},1\text{/}{3}^{2}\text{,\ldots }\) along with the function \(f(x)=1\text{/}{x}^{2}.\) From the graph we see that

\[\sum _{n=1}^{k}\frac{1}{{n}^{2}}=1+\frac{1}{{2}^{2}}+\frac{1}{{3}^{2}}+\text{\cdots }+\frac{1}{{k}^{2}}<1+{\int }_{1}^{k}\frac{1}{{x}^{2}}dx.\]

Therefore, for each \(k,\) the \(k\text{th}\) partial sum \({S}_{k}\) satisfies

\[{S}_{k}=\sum _{n=1}^{k}\frac{1}{{n}^{2}}<1+{\int }_{1}^{k}\frac{1}{{x}^{2}}dx=1-{\frac{1}{x}|}_{1}^{k}=1-\frac{1}{k}+1=2-\frac{1}{k}<2.\]

We conclude that the sequence of partial sums \(\{{S}_{k}\}\) is bounded. We also see that \(\{{S}_{k}\}\) is an increasing sequence:

\[{S}_{k}={S}_{k-1}+\frac{1}{{k}^{2}}\ \text{for}\ k\ge 2.\]\[{S}_{k}={a}_{1}+{a}_{2}+{a}_{3}+\text{\cdots }+{a}_{k}<{a}_{1}+{\int }_{1}^{k}f(x)dx<{a}_{1}+{\int }_{1}^{\infty }f(x)dx.\]\[{S}_{k}={a}_{1}+{a}_{2}+{a}_{3}+\text{\cdots }+{a}_{k}>{\int }_{1}^{k+1}f(x)dx.\]\[\sum _{n=1}^{\infty }{(\frac{1}{e})}^{n}=\frac{1}{e}+{(\frac{1}{e})}^{2}+{(\frac{1}{e})}^{3}+\text{\cdots }\]

Condensed — the full section is in OpenStax Calculus Volume 2.

The

The harmonic series \(\sum _{n=1}^{\infty }1\text{/}n\) and the series \(\sum _{n=1}^{\infty }1\text{/}{n}^{2}\) are both examples of a type of series called a p-series.

We know the p-series converges if \(p=2\) and diverges if \(p=1.\) What about other values of \(p\text{?}\) In general, it is difficult, if not impossible, to compute the exact value of most \(p\)-series. However, we can use the tests presented thus far to prove whether a \(p\)-series converges or diverges.

If \(p<0,\) then \(1\text{/}{n}^{p}\to \infty ,\) and if \(p=0,\) then \(1\text{/}{n}^{p}\to 1.\) Therefore, by the divergence test,

\[\sum _{n=1}^{\infty }1\text{/}{n}^{p}\ \text{diverges if}\ p\le 0.\]

If \(p>0,\) then \(f(x)=1\text{/}{x}^{p}\) is a positive, continuous, decreasing function. Therefore, for \(p>0,\) we use the integral test, comparing

\[\sum _{n=1}^{\infty }\frac{1}{{n}^{p}}\ \text{and}\ {\int }_{1}^{\infty }\frac{1}{{x}^{p}}dx.\]

We have already considered the case when \(p=1.\) Here we consider the case when \(p>0,p\ne 1.\) For this case,

\[{\int }_{1}^{\infty }\frac{1}{{x}^{p}}dx=\underset{b\to \infty }{\text{lim}}{\int }_{1}^{b}\frac{1}{{x}^{p}}dx=\underset{b\to \infty }{\text{lim}}\frac{1}{1-p}{x}^{1-p}{|}_{1}^{b}=\underset{b\to \infty }{\text{lim}}\frac{1}{1-p}[{b}^{1-p}-1].\]

Because

\[{b}^{1-p}\to 0\ \text{if}\ p>1\ \text{and}\ {b}^{1-p}\to \infty \ \text{if}\ p<1,\]

we conclude that

\[{\int }_{1}^{\infty }\frac{1}{{x}^{p}}dx=\{\begin{array}{l}\frac{1}{p-1}\ \text{if}\ p>1 \\ \infty \ \text{if}\ p\le 1\end{array}.\]\[\sum _{n=1}^{\infty }\frac{1}{{n}^{p}}\ \{\begin{array}{l}\text{converges if}\ p>1 \\ \text{diverges if}\ p\le 1\end{array}.\]

Example

Try it.

For each of the following series, determine whether it converges or diverges.

\(\sum _{n=1}^{\infty }\frac{1}{{n}^{4}}\)
\(\sum _{n=1}^{\infty }\frac{1}{{n}^{2\text{/}3}}\)

Solution

This is a p-series with \(p=4>1,\) so the series converges.
Since \(p=2\text{/}3<1,\) the series diverges.

Condensed — the full section is in OpenStax Calculus Volume 2.

Estimating the Value of a Series

Suppose we know that a series \(\sum _{n=1}^{\infty }{a}_{n}\) converges and we want to estimate the sum of that series. Certainly we can approximate that sum using any finite sum \(\sum _{n=1}^{N}{a}_{n}\) where \(N\) is any positive integer. The question we address here is, for a convergent series \(\sum _{n=1}^{\infty }{a}_{n},\) how good is the approximation \(\sum _{n=1}^{N}{a}_{n}\text{?}\) More specifically, if we let

\[{R}_{N}=\sum _{n=1}^{\infty }{a}_{n}-\sum _{n=1}^{N}{a}_{n}\]

be the remainder when the sum of an infinite series is approximated by the \(N\text{th}\) partial sum, how large is \({R}_{N}\text{?}\) For some types of series, we are able to use the ideas from the integral test to estimate \({R}_{N}.\)

We illustrate in . In particular, by representing the remainder \({R}_{N}={a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\text{\cdots }\) as the sum of areas of rectangles, we see that the area of those rectangles is bounded above by \({\int }_{N}^{\infty }f(x)dx\) and bounded below by \({\int }_{N+1}^{\infty }f(x)dx.\) In other words,

\[{R}_{N}={a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\text{\cdots }>{\int }_{N+1}^{\infty }f(x)dx\]

and

\[{R}_{N}={a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\text{\cdots }<{\int }_{N}^{\infty }f(x)dx.\]

We conclude that

\[{\int }_{N+1}^{\infty }f(x)dx<{R}_{N}<{\int }_{N}^{\infty }f(x)dx.\]

Since

\[\sum _{n=1}^{\infty }{a}_{n}={S}_{N}+{R}_{N},\]

where \({S}_{N}\) is the \(N\text{th}\) partial sum, we conclude that

\[{S}_{N}+{\int }_{N+1}^{\infty }f(x)dx<\sum _{n=1}^{\infty }{a}_{n}<{S}_{N}+{\int }_{N}^{\infty }f(x)dx.\]

Condensed — the full section is in OpenStax Calculus Volume 2.

Key Concepts

If \(\underset{n\to \infty }{\text{lim}}{a}_{n}\ne 0,\) then the series \(\sum _{n=1}^{\infty }{a}_{n}\) diverges.
If \(\underset{n\to \infty }{\text{lim}}{a}_{n}=0,\) the series \(\sum _{n=1}^{\infty }{a}_{n}\) may converge or diverge.
If \(\sum _{n=1}^{\infty }{a}_{n}\) is a series with positive terms \({a}_{n}\) and \(f\) is a continuous, decreasing function such that \(f(n)={a}_{n}\) for all positive integers \(n,\) then
\[\sum _{n=1}^{\infty }{a}_{n}\text{and}{\int }_{1}^{\infty }f(x)dx\]
either both converge or both diverge. Furthermore, if \(\sum _{n=1}^{\infty }{a}_{n}\) converges, then the \(N\text{th}\) partial sum approximation \({S}_{N}\) is accurate up to an error \({R}_{N}\) where \({\int }_{N+1}^{\infty }f(x)dx<{R}_{N}<{\int }_{N}^{\infty }f(x)dx.\)
The p-series \(\sum _{n=1}^{\infty }1\text{/}{n}^{p}\) converges if \(p>1\) and diverges if \(p\le 1.\)

Key Equations

Divergence test	\(\text{If}\ {a}_{n}↛0\ \text{as}\ n\to \infty ,\sum _{n=1}^{\infty }{a}_{n}\ \text{diverges}.\)
p-series	\(\sum _{n=1}^{\infty }\frac{1}{{n}^{p}}\{\begin{array}{l}\text{converges if}\ p>1 \\ \text{diverges if}\ p\le 1\end{array}\)
Remainder estimate from the integral test	\({\int }_{N+1}^{\infty }f(x)dx<{R}_{N}<{\int }_{N}^{\infty }f(x)dx\)

The Divergence and Integral Tests

For each of the following series, if the divergence test applies, either state that \(\underset{n\to \infty }{\text{lim}}{a}_{n}\) does not exist or find \(\underset{n\to \infty }{\text{lim}}{a}_{n}.\) If the divergence test does not apply, state why.

State whether the given \(p\)-series converges.

Use the integral test to determine whether the following sums converge.

Express the following sums as \(p\)-series and determine whether each converges.

Use the estimate \({R}_{N}\le {\int }_{N}^{\infty }f(t)dt\) to find a bound for the remainder \({R}_{N}=\sum _{n=1}^{\infty }{a}_{n}-\sum _{n=1}^{N}{a}_{n}\) where \({a}_{n}=f(n).\)

[T] Find the minimum value of \(N\) such that the remainder estimate \({\int }_{N+1}^{\infty }f<{R}_{N}<{\int }_{N}^{\infty }f\) guarantees that \(\sum _{n=1}^{N}{a}_{n}\) estimates \(\sum _{n=1}^{\infty }{a}_{n},\) accurate to within the given error.

In the following exercises, find a value of \(N\) such that \({R}_{N}\) is smaller than the desired error. Compute the corresponding sum \(\sum _{n=1}^{N}{a}_{n}\) and compare it to the given estimate of the infinite series.

Condensed — the full section is in OpenStax Calculus Volume 2.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

For each of the following series, apply the divergence test. If the divergence test proves that the series diverges, state so. Otherwise, indicate that the divergence test is inconclusive.
1. \(\sum _{n=1}^{\infty }\frac{n}{3n-1}\)
2. \(\sum _{n=1}^{\infty }\frac{1}{{n}^{3}}\)
3. \(\sum _{n=1}^{\infty }{e}^{1\text{/}{n}^{2}}\)
Kuratidza mhinduro
1. Since \(n\text{/}(3n-1)\to 1\text{/}3\ne 0,\) by the divergence test, we can conclude that
  \[\sum _{n=1}^{\infty }\frac{n}{3n-1}\]
  diverges.
2. Since \(1\text{/}{n}^{3}\to 0,\) the divergence test is inconclusive.
3. Since \({e}^{1\text{/}{n}^{2}}\to 1\ne 0,\) by the divergence test, the series
  \[\sum _{n=1}^{\infty }{e}^{1\text{/}{n}^{2}}\]
  diverges.
What does the divergence test tell us about the series \(\sum _{n=1}^{\infty }\text{cos}(1\text{/}{n}^{2})\text{?}\)

Kuratidza mhinduro

The series diverges.
For each of the following series, use the integral test to determine whether the series converges or diverges. Assume that all conditions for the integral test are met.
1. \(\sum _{n=1}^{\infty }1\text{/}{n}^{3}\)
2. \(\sum _{n=1}^{\infty }1\text{/}\sqrt{2n-1}\)
Kuratidza mhinduro
1. Compare
  \[\sum _{n=1}^{\infty }\frac{1}{{n}^{3}}\ \text{and}\ {\int }_{1}^{\infty }\frac{1}{{x}^{3}}dx.\]
  We have
  \[{\int }_{1}^{\infty }\frac{1}{{x}^{3}}dx=\underset{b\to \infty }{\text{lim}}{\int }_{1}^{b}\frac{1}{{x}^{3}}dx=\underset{b\to \infty }{\text{lim}}[-\frac{1}{2{x}^{2}}{|}_{1}^{b}]=\underset{b\to \infty }{\text{lim}}[-\frac{1}{2{b}^{2}}+\frac{1}{2}]=\frac{1}{2}.\]
  Thus the integral \({\int }_{1}^{\infty }1\text{/}{x}^{3}dx\) converges, and therefore so does the series
  \[\sum _{n=1}^{\infty }\frac{1}{{n}^{3}}.\]
2. Compare
  \[\sum _{n=1}^{\infty }\frac{1}{\sqrt{2n-1}}\ \text{and}\ {\int }_{1}^{\infty }\frac{1}{\sqrt{2x-1}}dx.\]
  Since
  \[\begin{array}{ll}{\int }_{1}^{\infty }\frac{1}{\sqrt{2x-1}}dx & =\underset{b\to \infty }{\text{lim}}{\int }_{1}^{b}\frac{1}{\sqrt{2x-1}}dx=\underset{b\to \infty }{\text{lim}}\sqrt{2x-1}{|}_{1}^{b} \\ & =\underset{b\to \infty }{\text{lim}}[\sqrt{2b-1}-1]=\infty ,\end{array}\]
  the integral \({\int }_{1}^{\infty }1\text{/}\sqrt{2x-1}\ dx\) diverges, and therefore
  \[\sum _{n=1}^{\infty }\frac{1}{\sqrt{2n-1}}\]
  diverges.
Use the integral test to determine whether the series \(\sum _{n=1}^{\infty }\frac{n}{3{n}^{2}+1}\) converges or diverges.

Kuratidza mhinduro

The series diverges.
For each of the following series, determine whether it converges or diverges.
1. \(\sum _{n=1}^{\infty }\frac{1}{{n}^{4}}\)
2. \(\sum _{n=1}^{\infty }\frac{1}{{n}^{2\text{/}3}}\)
Kuratidza mhinduro
1. This is a p-series with \(p=4>1,\) so the series converges.
2. Since \(p=2\text{/}3<1,\) the series diverges.
Does the series \(\sum _{n=1}^{\infty }\frac{1}{{n}^{5\text{/}4}}\) converge or diverge?

Kuratidza mhinduro

The series converges.
Consider the series \(\sum _{n=1}^{\infty }1\text{/}{n}^{3}.\)
1. Calculate \({S}_{10}=\sum _{n=1}^{10}1\text{/}{n}^{3}\) and estimate the error.
2. Determine the least value of \(N\) necessary such that \({S}_{N}\) will estimate \(\sum _{n=1}^{\infty }1\text{/}{n}^{3}\) to within \(0.001.\)
Kuratidza mhinduro
1. Using a calculating utility, we have
  \[{S}_{10}=1+\frac{1}{{2}^{3}}+\frac{1}{{3}^{3}}+\frac{1}{{4}^{3}}+\text{\cdots }+\frac{1}{{10}^{3}}\approx 1.19753.\]
  By the remainder estimate, we know
  \[{R}_{N}<{\int }_{N}^{\infty }\frac{1}{{x}^{3}}dx.\]
  We have
  \[{\int }_{10}^{\infty }\frac{1}{{x}^{3}}dx=\underset{b\to \infty }{\text{lim}}{\int }_{10}^{b}\frac{1}{{x}^{3}}dx=\underset{b\to \infty }{\text{lim}}[-\frac{1}{2{x}^{2}}{]}_{N}^{b}=\underset{b\to \infty }{\text{lim}}[-\frac{1}{2{b}^{2}}+\frac{1}{2{N}^{2}}]=\frac{1}{2{N}^{2}}.\]
  Therefore, the error is \({R}_{10}<1\text{/}2{(10)}^{2}=0.005.\)
2. Find \(N\) such that \({R}_{N}<0.001.\) In part a. we showed that \({R}_{N}<1\text{/}\left(2{N}^{2}\right).\) Therefore, the remainder \({R}_{N}<0.001\) as long as \(1\text{/}\left(2{N}^{2}\right)<0.001.\) That is, we need \(2{N}^{2}>1000.\) Solving this inequality for \(N,\) we see that we need \(N>22.36.\) To ensure that the remainder is within the desired amount, we need to round up to the nearest integer. Therefore, the minimum necessary value is \(N=23.\)
For \(\sum _{n=1}^{\infty }\frac{1}{{n}^{4}},\) calculate \({S}_{5}\) and estimate the error \({R}_{5}.\)

Kuratidza mhinduro

\({S}_{5}\approx 1.09035,\) \({R}_{5}<0.00267\)
\({a}_{n}=\frac{n}{n+2}\)
\({a}_{n}=\frac{n}{5{n}^{2}-3}\)

Kuratidza mhinduro

\(\underset{n\to \infty }{\text{lim}}{a}_{n}=0.\) Divergence test does not apply.
\({a}_{n}=\frac{n}{\sqrt{3{n}^{2}+2n+1}}\)
\({a}_{n}=\frac{(2n+1)(n-1)}{{(n+1)}^{2}}\)

Kuratidza mhinduro

\(\underset{n\to \infty }{\text{lim}}{a}_{n}=2.\) Series diverges.
\({a}_{n}=\frac{{(2n+1)}^{2n}}{{(3{n}^{2}+1)}^{n}}\)
\({a}_{n}=\frac{{2}^{n}}{{3}^{n\text{/}2}}\)

Kuratidza mhinduro

\(\underset{n\to \infty }{\text{lim}}{a}_{n}=\infty\) (does not exist). Series diverges.
\({a}_{n}=\frac{{2}^{n}+{3}^{n}}{{10}^{n\text{/}2}}\)
\({a}_{n}={e}^{-2\text{/}n}\)

Kuratidza mhinduro

\(\underset{n\to \infty }{\text{lim}}{a}_{n}=1.\) Series diverges.
\({a}_{n}=\text{cos}\ n\)
\({a}_{n}=\text{tan}\ n\)

Kuratidza mhinduro

\(\underset{n\to \infty }{\text{lim}}{a}_{n}\) does not exist. Series diverges.
\({a}_{n}=\frac{1-{\text{cos}}^{2}(1\text{/}n)}{{\text{sin}}^{2}(2\text{/}n)}\)
\({a}_{n}={(1-\frac{1}{n})}^{2n}\)

Kuratidza mhinduro

\(\underset{n\to \infty }{\text{lim}}{a}_{n}=1\text{/}{e}^{2}.\) Series diverges.
\({a}_{n}=\frac{\text{ln}\ n}{n}\)
\({a}_{n}=\frac{{(\text{ln}\ n)}^{2}}{\sqrt{n}}\)

Kuratidza mhinduro

\(\underset{n\to \infty }{\text{lim}}{a}_{n}=0.\) Divergence test does not apply.
\(\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}\)
\(\sum _{n=1}^{\infty }\frac{1}{n\sqrt{n}}\)

Kuratidza mhinduro

Series converges, \(p>1.\)
\(\sum _{n=1}^{\infty }\frac{1}{\sqrt[3]{{n}^{2}}}\)
\(\sum _{n=1}^{\infty }\frac{1}{\sqrt[3]{{n}^{4}}}\)

Kuratidza mhinduro

Series converges, \(p=4\text{/}3>1.\)
\(\sum _{n=1}^{\infty }\frac{{n}^{e}}{{n}^{\pi }}\)
\(\sum _{n=1}^{\infty }\frac{{n}^{\pi }}{{n}^{2e}}\)

Kuratidza mhinduro

Series converges, \(p=2e-\pi >1.\)
\(\sum _{n=1}^{\infty }\frac{1}{\sqrt{n+5}}\)
\(\sum _{n=1}^{\infty }\frac{1}{\sqrt[3]{n+5}}\)

Kuratidza mhinduro

Series diverges by comparison with \({\int }_{1}^{\infty }\frac{dx}{{(x+5)}^{1\text{/}3}}.\)
\(\sum _{n=2}^{\infty }\frac{1}{n\ \text{ln}\ n}\)
\(\sum _{n=1}^{\infty }\frac{n}{1+{n}^{2}}\)

Kuratidza mhinduro

Series diverges by comparison with \({\int }_{1}^{\infty }\frac{x}{1+{x}^{2}}dx.\)
\(\sum _{n=1}^{\infty }\frac{{e}^{n}}{1+{e}^{2n}}\)
\(\sum _{n=1}^{\infty }\frac{2n}{1+{n}^{4}}\)

Kuratidza mhinduro

Series converges by comparison with \({\int }_{1}^{\infty }\frac{2x}{1+{x}^{4}}dx.\)
\(\sum _{n=2}^{\infty }\frac{1}{n\ {\text{ln}}^{2}n}\)
\(\sum _{n=1}^{\infty }{2}^{\text{-}\text{ln}\ n}\) (Hint: \({2}^{\text{-}\text{ln}\ n}=1\text{/}{n}^{\text{ln}\ 2}\).)

Kuratidza mhinduro

\({2}^{\text{-}\text{ln}\ n}=1\text{/}{n}^{\text{ln}\ 2}.\) Since \(\text{ln}\ 2<1,\) diverges by \(p\)-series.
\(\sum _{n=1}^{\infty }{3}^{\text{-}\text{ln}\ n}\) (Hint: \({3}^{\text{-}\text{ln}\ n}=1\text{/}{n}^{\text{ln}\ 3}\).)
\(\sum _{n=1}^{\infty }n{2}^{-2\ \text{ln}\ n}\)

Kuratidza mhinduro

\({2}^{-2\ \text{ln}\ n}=1\text{/}{n}^{2\ \text{ln}\ 2}.\) Since \(2\ \text{ln}\ 2-1<1,\) diverges by \(p\)-series.
\(\sum _{n=1}^{\infty }n{3}^{-2\ \text{ln}\ n}\)
\(\sum _{n=1}^{1000}\frac{1}{{n}^{2}}\)

Kuratidza mhinduro

\({R}_{1000}\le {\int }_{1000}^{\infty }\frac{dt}{{t}^{2}}=-\frac{1}{t}{|}_{1000}^{\infty }=0.001\)

Symbols used here

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\approx

approximately equal
Equal to the precision shown, not exactly.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\neq

not equal
The two sides are different.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

Euler's number
2.71828…, the base whose exponential is its own derivative.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: The Divergence and Integral Tests

Use the divergence test to demonstrate that a series diverges.
Use the integral test to determine the convergence of a series.
Estimate the value of a series by finding bounds on its remainder term.
Since
Since
Since
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Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

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Parts of this page are adapted from OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests