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Find all possible values of the positive constant k such that the series

$\sum_{n=1}^{\infty}\ln\left(1+\frac{1}{n^k}\right)$

is convergent.


Definitely not root test. Tried ratio test, $L = 1$ which is not conclusive.

Tried integral test but the integral looks too hideous to evaluate.

Any other suggestions?


How do I come up with the series to compare with? Keep practising?

For this question, in my mind, I would compare it with another logarithm and not even think about Riemann series.

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    @Didier: I see, sorry, I overlooked that we only need it in $(0,1)$.2011-11-13

2 Answers 2

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I will give some hints. For $0, $\log(1+x)\leq x$. Thus $\log (1+n^{-k}) \leq n^{-k}$ and the comparison test can be used.

This leaves $k \leq 1$. Do $k=1$: I suggest writing $1+1/n = (n+1)/n$ then writing $\log( (n+1)/n)$ as $\log(n+1)-\log(n)$. Now you can find the partial sums explicitly.

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You can use the integral test. For $ \int \ln(1+x^{-k})\, dx, $ use integration by parts with $dv=1\,dx$. This gives $ \int \underbrace{\ln(1+x^{-k})}_u\underbrace{1\cdot dx\vphantom{(}}_{dv} = \underbrace{\ln(1+x^{-k})}_u \cdot \underbrace{x\vphantom{(}}_v- \int \underbrace{\vphantom{k\over x}x}_v\cdot \underbrace{{-kx^{-k-1}\over 1+x^{-k}}\,dx }_{du} . $ A bit of simplification gives: $ \int \ln(1+x^{-k})\, dx=\underbrace{\vphantom{\int} x\ln(1+x^{-k})}_{=A} +\underbrace{\int { k \over x^k+1}\,dx}_{=B} $

Note that $A$ and the integrand in $B$ are positive for positive $x$.

The integrand in $B$ satisfies $ {k\over 2x^k}<{k\over 1+x^k}\le {k\over x^k}.$ Thus $\int_1^\infty {k\over x^k+1}\,dx$ converges if and only if $k>1$. This implies the series diverges for $k\le 1$.

For $k>1$, $\eqalign{\lim_{x\rightarrow\infty} A &=\lim_{x\rightarrow\infty} {\ln(1+x^{-k})\over 1/x}\cr &=\lim_{x\rightarrow\infty} {-kx^{-k-1}/(1+x^{-k}) \over -1/x^2}\cr &= \lim_{x\rightarrow\infty} {kx^{-k+1}\over1+x^{-k}}\cr &= 0. }$ This, together with the previous observation that $\int_1^\infty {k\over x^k+1}\,dx$ converges for $k>1$ implies that the series converges for $k>1$.