maths.free › Calculus › 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.

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

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

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

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

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

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:

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

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.

  1. \(\sum _{n=1}^{\infty }\frac{1}{{n}^{4}}\)
  2. \(\sum _{n=1}^{\infty }\frac{1}{{n}^{2\text{/}3}}\)
Solution
  1. This is a p-series with \(p=4>1,\) so the series converges.
  2. 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 \({R}_{N}={a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\cdots\) 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,

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

and

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

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.

Haddaad Ma jiro xisaabis oo ku nool midkan, laakiin qaybaha ka mid ah waa la xisaabi karaa. Ku raaxayso mid ka hooseeya, ama ku qor adiga kuu gaar ah.

Tababar (40)

Ku day mid kasta oo ka mid ah oo ku saabsan buugaag hore. Muuji jawaabta si aad u hubiso; kuwa la xaqiijiyay waxaa la furi karaa in xalka ah tallaabo kasta.

  1. 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. \(\sum _{n=1}^{\infty }{e}^{1\text{/}{n}^{2}}\)
    Jawaabta muuji
    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.

  2. What does the divergence test tell us about the series \(\sum _{n=1}^{\infty }\text{cos}(1\text{/}{n}^{2})\text{?}\)

    Jawaabta muuji

    The series diverges.

  3. 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. \(\sum _{n=1}^{\infty }1\text{/}\sqrt{2n-1}\)
    Jawaabta muuji
    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.

  4. Use the integral test to determine whether the series \(\sum _{n=1}^{\infty }\frac{n}{3{n}^{2}+1}\) converges or diverges.

    Jawaabta muuji

    The series diverges.

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

    1. \(\sum _{n=1}^{\infty }\frac{1}{{n}^{4}}\)
    2. \(\sum _{n=1}^{\infty }\frac{1}{{n}^{2\text{/}3}}\)
    Jawaabta muuji
    1. This is a p-series with \(p=4>1,\) so the series converges.
    2. Since \(p=2\text{/}3<1,\) the series diverges.

  6. Does the series \(\sum _{n=1}^{\infty }\frac{1}{{n}^{5\text{/}4}}\) converge or diverge?

    Jawaabta muuji

    The series converges.

  7. 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 \(0.001.\)
    Jawaabta muuji
    1. Using a calculating utility, we have
      \[{S}_{10}=1+\frac{1}{{2}^{3}}+\frac{1}{{3}^{3}}+\frac{1}{{4}^{3}}+\cdots +\frac{1}{{10}^{3}}\approx 1.19753.\]
      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 \(N=23.\)

  8. For \(\sum _{n=1}^{\infty }\frac{1}{{n}^{4}},\) calculate \({S}_{5}\) and estimate the error \({R}_{5}.\)

    Jawaabta muuji

    \({S}_{5}\approx 1.09035,\) \({R}_{5}<0.00267\)

  9. \({a}_{n}=\frac{n}{n+2}\)

  10. \({a}_{n}=\frac{n}{5{n}^{2}-3}\)

    Jawaabta muuji

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

  11. \({a}_{n}=\frac{n}{\sqrt{3{n}^{2}+2n+1}}\)

  12. \({a}_{n}=\frac{(2n+1)(n-1)}{{(n+1)}^{2}}\)

    Jawaabta muuji

    \(\underset{n\to \infty }{\text{lim}}{a}_{n}=2.\) Series diverges.

  13. \({a}_{n}=\frac{{(2n+1)}^{2n}}{{(3{n}^{2}+1)}^{n}}\)

  14. \({a}_{n}=\frac{{2}^{n}}{{3}^{n\text{/}2}}\)

    Jawaabta muuji

    \(\underset{n\to \infty }{\text{lim}}{a}_{n}=\infty\) (does not exist). Series diverges.

  15. \({a}_{n}=\frac{{2}^{n}+{3}^{n}}{{10}^{n\text{/}2}}\)

  16. \({a}_{n}={e}^{-2\text{/}n}\)

    Jawaabta muuji

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

  17. \({a}_{n}=\text{cos}\ n\)

  18. \({a}_{n}=\text{tan}\ n\)

    Jawaabta muuji

    \(\underset{n\to \infty }{\text{lim}}{a}_{n}\) does not exist. Series diverges.

  19. \({a}_{n}=\frac{1-{\text{cos}}^{2}(1\text{/}n)}{{\text{sin}}^{2}(2\text{/}n)}\)

  20. \({a}_{n}={(1-\frac{1}{n})}^{2n}\)

    Jawaabta muuji

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

  21. \({a}_{n}=\frac{\text{ln}\ n}{n}\)

  22. \({a}_{n}=\frac{{(\text{ln}\ n)}^{2}}{\sqrt{n}}\)

    Jawaabta muuji

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

  23. \(\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}\)

  24. \(\sum _{n=1}^{\infty }\frac{1}{n\sqrt{n}}\)

    Jawaabta muuji

    Series converges, \(p>1.\)

  25. \(\sum _{n=1}^{\infty }\frac{1}{\sqrt[3]{{n}^{2}}}\)

  26. \(\sum _{n=1}^{\infty }\frac{1}{\sqrt[3]{{n}^{4}}}\)

    Jawaabta muuji

    Series converges, \(p=4\text{/}3>1.\)

  27. \(\sum _{n=1}^{\infty }\frac{{n}^{e}}{{n}^{\pi }}\)

  28. \(\sum _{n=1}^{\infty }\frac{{n}^{\pi }}{{n}^{2e}}\)

    Jawaabta muuji

    Series converges, \(p=2e-\pi >1.\)

  29. \(\sum _{n=1}^{\infty }\frac{1}{\sqrt{n+5}}\)

  30. \(\sum _{n=1}^{\infty }\frac{1}{\sqrt[3]{n+5}}\)

    Jawaabta muuji

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

  31. \(\sum _{n=2}^{\infty }\frac{1}{n\ \text{ln}\ n}\)

  32. \(\sum _{n=1}^{\infty }\frac{n}{1+{n}^{2}}\)

    Jawaabta muuji

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

  33. \(\sum _{n=1}^{\infty }\frac{{e}^{n}}{1+{e}^{2n}}\)

  34. \(\sum _{n=1}^{\infty }\frac{2n}{1+{n}^{4}}\)

    Jawaabta muuji

    Series converges by comparison with \({\int }_{1}^{\infty }\frac{2x}{1+{x}^{4}}dx.\)

  35. \(\sum _{n=2}^{\infty }\frac{1}{n\ {\text{ln}}^{2}n}\)

  36. \(\sum _{n=1}^{\infty }{2}^{\text{-}\text{ln}\ n}\) (Hint: \({2}^{\text{-}\text{ln}\ n}=1\text{/}{n}^{\text{ln}\ 2}\).)

    Jawaabta muuji

    \({2}^{\text{-}\text{ln}\ n}=1\text{/}{n}^{\text{ln}\ 2}.\) Since \(\text{ln}\ 2<1,\) diverges by \(p\)-series.

  37. \(\sum _{n=1}^{\infty }{3}^{\text{-}\text{ln}\ n}\) (Hint: \({3}^{\text{-}\text{ln}\ n}=1\text{/}{n}^{\text{ln}\ 3}\).)

  38. \(\sum _{n=1}^{\infty }n{2}^{-2\ \text{ln}\ n}\)

    Jawaabta muuji

    \({2}^{-2\ \text{ln}\ n}=1\text{/}{n}^{2\ \text{ln}\ 2}.\) Since \(2\ \text{ln}\ 2-1<1,\) diverges by \(p\)-series.

  39. \(\sum _{n=1}^{\infty }n{3}^{-2\ \text{ln}\ n}\)

  40. \(\sum _{n=1}^{1000}\frac{1}{{n}^{2}}\)

    Jawaabta muuji

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

Ku hay shaqadaada

A xisaab bilaash ah oo ku daray qoraalada waxbarasho kasta, diiwaanka waxa aad dhamaatay, dhibaatooyinka aad xal u tahay meel kaliya, iyo macallinka aad weydiisan kartaa oo ku saabsan bogga. Xisaabta oo keliya waa u furan yahay qof kasta, la diiwaan geliyey ama aan.

Ku soo biir Ku soo gal

Calaamadaha la isticmaalay halkan

Taab calaamad kasta oo loogu talagalay qeexidda buuxda, sawir, iyo waxa xarfaha oo dhan ka mid ah ka dhigan tahay.

Sida loo: The Divergence and Integral Tests

  1. Use the divergence test to demonstrate that a series diverges.
  2. Use the integral test to determine the convergence of a series.
  3. Estimate the value of a series by finding bounds on its remainder term.
  4. Since
  5. Since
  6. Since
  7. Compare
  8. Compare

Su'aalaha dadku waydiiyaan

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.

Qaybaha ee boggaan waxaa la beddelay ka OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed iyo dib-oo halkan lagu sharaxay; khaladaadka waa inoo. Sababtoo ah asalka waxaa loo oggolaaday in aan loo isticmaalin ganacsi, fasalkan waa bilaash, waxba ma iibiyo, iyo waxaa lagu wadaagayaa liisan la mid ah.

In ka badan Calculus

LimitsDerivativesIntegralsDefinite integralsTaylor seriesSeries and sumsMaxima and minimaThe chain ruleImplicit differentiationRelated rates and optimisationIntegration techniques: substitution, parts, partial fractionsApplications of integration: area, volume, arc lengthInfinite series and convergence tests