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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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Compare the general term $\ln(1+n^{-k})$ to a power of $n$ and use the known results about Riemann series (link in French, apparently the page does not exist in English). –  Did Nov 13 '11 at 10:22
    
Use $\backslash$ln. –  Did Nov 13 '11 at 10:22
    
@Didier: That doesn't work for $k\le1$, since we'd need the comparison to go the other way to demonstrate divergence. –  joriki Nov 13 '11 at 10:32
    
@joriki, sorry? $\ln(1+x)\geqslant \frac12x$ for every $x$ in $(0,1)$. –  Did Nov 13 '11 at 10:36
    
@Didier: I see, sorry, I overlooked that we only need it in $(0,1)$. –  joriki Nov 13 '11 at 10:40

2 Answers 2

I will give some hints. For $0<x<1$, $\log(1+x)\leq x$. 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 $k>1$.

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