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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.

Exemplo resolvido · does 1/n^2 for n = 1 to oo converge

Calculus · Series-And-Sums

Convergence of n^(-2)

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

Passo a passo

  1. \sum_{n=1}^{\infty} \frac{1}{n^{2}}

    Does this series converge?

  2. \lim_{n\to\infty} \frac{1}{n^{2}} = 0

    Divergence test: if the terms do not go to 0 the series cannot converge.

  3. \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.

  4. = \frac{\pi^{2}}{6}

    It converges, and the CAS knows the sum.

Revelar a resposta
\text{converges to } \frac{\pi^{2}}{6}

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Como: Convergence of series

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

Perguntas que as pessoas fazem

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.

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