0
$\begingroup$

The sum is $\sum_{i=1}^{\infty} \frac{a^n}{n}$, where $0<a<1$. One easy upper bound is $\frac{a}{1-a}$. Are there any tighter upper bound available?

$\endgroup$
  • 1
    $\begingroup$ $\displaystyle \sum_{n=1}^{\infty} \frac{a^n}{n} = -\ln(1-a)$ $\endgroup$ – DHMO Apr 13 '17 at 17:31
  • 1
    $\begingroup$ Let $\displaystyle f(a) = \sum_{n=1}^{\infty} \frac{a^n}{n}$. Differentiate both sides. Apply geometric series. Integrate both sides. $\endgroup$ – DHMO Apr 13 '17 at 17:32
  • $\begingroup$ Clearly the partial sums $\sum^N_{n=1} \frac{a^n}n$ are increasing (as $N$ increases) since we are simply adding more positive terms (when $0<a<1$). Thus they converge monotonically to their limit: $-\ln(1-a)$. $\endgroup$ – User8128 Apr 13 '17 at 17:34
  • $\begingroup$ @DHMO Thank you very much! $\endgroup$ – user125056 Apr 13 '17 at 17:36
  • $\begingroup$ @DHMO Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. $\endgroup$ – kingW3 Apr 18 '17 at 16:44
1
$\begingroup$

$$\begin{array}{rcl} f(a) &=& \displaystyle \sum_{n=1}^{\infty} \frac{a^n}{n} \\ f'(a) &=& \displaystyle \sum_{n=1}^{\infty} \frac{na^{n-1}}{n} \\ &=& \displaystyle \sum_{n=1}^{\infty} a^{n-1} \\ &=& \displaystyle \sum_{n=0}^{\infty} a^n \\ &=& \dfrac1{1-a} \\ \displaystyle \int_0^a f'(a) \ \mathrm da &=& \displaystyle \int_0^a \dfrac1{1-a} \ \mathrm da \\ f(a) - f(0) &=& -\ln(1-a) - 0 \\ f(a) - 0 &=& -\ln(1-a) - 0 \\ f(a) &=& -\ln(1-a)\\ \end{array}$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.