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Convergence of series

Divergence, ratio, comparison and integral tests, and the harmonic series.

An infinite series converges when its partial sums approach a limit. The terms going to zero is necessary but not enough. The harmonic series 1 + ½ + ⅓ + … diverges even though its terms vanish. The ratio test settles most series with factorials or powers; comparison and the integral test handle the p-series family.

Àwọn ààtò àwọn ìṣàmúlò-ètò · does 1/n^2 for n = 1 to oo converge

Calculus · Series-And-Sums

Convergence of n^(-2)

\frac{1}{n^{2}}

Àwọn ìṣàmúlò-ètò

\sum_{n=1}^{\infty} \frac{1}{n^{2}}
Does this series converge?
\lim_{n\to\infty} \frac{1}{n^{2}} = 0
Divergence test: if the terms do not go to 0 the series cannot converge.
\lim \left|\frac{a_{k+1}}{a_k}\right| = \lim \left|{\frac{n^{2}}{\left(n + 1\right)^{2}}}\right| = 1
Ratio test: < 1 converges, > 1 diverges, = 1 says nothing.
= \frac{\pi^{2}}{6}
It converges, and the CAS knows the sum.

Fi àwọn àgbèwọlé hàn

\text{converges to } \frac{\pi^{2}}{6}

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Bii o ṣe le: Convergence of series

Does this series converge?
Divergence test: if the terms do not go to 0 the series cannot converge.
Ratio test: < 1 converges, > 1 diverges, = 1 says nothing.
It converges, and the CAS knows the sum.

Àwọn Àtòjọ-ẹ̀yàn

What is the ε, δ definition actually saying?

That you can make the output as close to the limit as anyone demands (within ε) by keeping the input close enough (within δ). It replaces "approaches" with a challenge-and-response that can be checked.

Why does the harmonic series diverge when its terms go to zero?

Because the terms shrink too slowly: group them as 1/3 + 1/4 > 1/2, 1/5 + … + 1/8 > 1/2, and so on: infinitely many halves.

Diẹ̀ nínú Real Analysis

Sequences and their limits Improper integrals Taylor approximation and error Functions of a complex variable Continuity and differentiability, rigorously Sequences of functions and uniform convergence