Series Convergence

[honest spade]

Any hint on how to determine wether this series converges or not

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[honest spade]

Cannot use integrals or power series

[robust jungle]

Détermine an equivalent of the main term of the series

to something familiar

ln(1+1/n)~?

[honest spade]

wdym?

[robust jungle]

if $(a_n)$ and $(b_n)$ are positive real sequences such that $a_n \sim_{n \to +\infty} b_n$ then $\sum_{n \geq 0} a_n$ converges if and only if $\sum_{n \geq 0} b_n$ converges

[proper folio]

R

[robust jungle]

so here you can find an equivalent of the main term of your series that you know the convergence of the series

(Like Riemann series)

[honest spade]

what does the equivalence symbol stand for?

[robust jungle]

Never mind that then use the fact that ln(1+x)<=x for x>0

then try comparing to a Riemann series

[honest spade]

$\ln(1+\frac{2}{n-1}) < \frac{2}{n-1}$

[proper folio]

Halex

[honest spade]

\frac{2}{n-1} is positive since n >= 2

[honest spade]

This implies that $$\frac{1}{\sqrt{n}} \cdot \ln(1+\frac{2}{n-1}) < \frac{2}{\sqrt{n} \cdot (n-1)} $$

[proper folio]

Halex

[honest spade]

And the Series in the r.h.s converges, tho it is still hard to determine it converges

[robust jungle]

Well the series 1/sqrt(n) n converges it’s the sum of 1/n^(3/2) 3/2>1

[honest spade]

but cannot split the denominator

[honest spade]

+close

[simple tide]

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