2
$\begingroup$

Evaluate the sum $S=\sum_{k=2}^{\infty} \frac{\zeta (k)-1}{k+1}$, where $\zeta (s)$ denotes the Riemann zeta function.

The sum is equal to $\sum_{k=2}^{\infty} \sum_{n=2}^{\infty} \frac{1}{(k+1)n^k}$, then switching the order (since the summand converges uniformly) gives $$S=\sum_{n=2}^{\infty} \sum_{k=2}^{\infty} \frac{1}{(k+1)n^k}$$.

For the inner sum, I tried to evaluate it by considering $\sum_{k=0}^{\infty} x^k = \frac{1}{1-x}$ for $|x|<1$. Integrating w.r.t $x$ and then divided both sides by $x$ gives

$$\sum_{k=2}^{\infty} \frac{x^k}{k+1}= \frac{-1}{x}\ln |1-x|-1-\frac{x}{2}$$. Then put $x=1/n$ for $n=2,3,\cdots$. But then I dont how to evaluate the outer sum, please helps.

$\endgroup$
2
  • $\begingroup$ The outer sum you are trying to evaluate is $$\sum_{n=2}^\infty -n \ln \left|1-\frac1n\right| - 1 - \frac1{2n} = \sum_{n=2}^\infty -n \ln \left(1-\frac1m\right) - 1 - \frac1{2n} \stackrel{\text{W.A.}}= \frac12 (3-\gamma-\ln(2\pi))$$ right? Wolfram Alpha link $\endgroup$
    – AlexR
    Commented Aug 23, 2014 at 13:17
  • $\begingroup$ The $\frac{3-\gamma}2$ part is kind of trivial to obtain (just take the partial sums, make the sum $\sum_{n=1}^N \frac 1n - \log N$ appear and isolate constants) but I have no idea about how to show the remaining term goes to $(1/2) \log(2\pi)$. For me the remaining term looks like this : $$ \left( (N-1/2) \log (N) - N - \sum_{n=2}^{N-1} \log(n) \right) \longrightarrow \frac 12 \log(2\pi). $$ $\endgroup$ Commented Aug 23, 2014 at 13:45

1 Answer 1

1
$\begingroup$

Remember Stirling's formula

$$\sum_{m=1}^n \log m = \log (n!) = \left(n+\tfrac{1}{2}\right)\log n - n + \tfrac{1}{2}\log (2\pi) + O\left(\tfrac{1}{n}\right).$$

Then we have

$$\begin{align} \sum_{n=2}^K \left(-n\log\left(1 - \tfrac{1}{n}\right) - 1 - \tfrac{1}{2n}\right) &= 1-K +\tfrac{1}{2} - \tfrac{1}{2}H_K - \sum_{n=2}^K n\log(n-1) + \sum_{n=2}^K n\log n\\ &= \tfrac{3}{2} - K - \tfrac{1}{2}H_K - \sum_{n=2}^{K-1}\log n + K\log K\\ &= \left(\left(K+\tfrac{1}{2}\right)\log K - K + \tfrac{1}{2}\log (2\pi) - \sum_{n=1}^K\log n\right)\\ &\qquad + \tfrac{3}{2}+ \tfrac{1}{2}\underbrace{\left(\log K - H_K\right)}_{-\gamma + O(K^{-1})} - \tfrac{1}{2}\log (2\pi)\\ &= \tfrac{3}{2} - \tfrac{1}{2}\gamma - \tfrac{1}{2}\log (2\pi) + O\left(\tfrac{1}{K}\right), \end{align}$$

where $H_K$ is the $K$-th harmonic number.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .