Find the following sum : $\sum_{n=1}^{\infty} \frac{\ln n }{n(n- \ln n)}$
I couldn't find any way to solve.
I just want a hint, how to initiate?
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1Have you tried partial fractions? By the way the sum does not seem to have a closed form. – Toby Mak Oct 07 '19 at 10:47
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@ClementYung Using '$$' makes the post easier to read. – Toby Mak Oct 07 '19 at 10:54
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An answer from this post suggests that $\sum_{n=1}^{\infty} \frac{\ln n}{n^2} = - \zeta' (2)$. – Toby Mak Oct 07 '19 at 10:58
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2What convinces you to believe that this sum has a closed form? Notice that your sum can be rephrased as $$\sum_{n=1}^{\infty} \frac{\log n}{n(n-\log n)} = \sum_{k=1}^{\infty} (-1)^k \zeta^{(k)}(k+1). $$ – Sangchul Lee Oct 07 '19 at 10:59
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Did your friend give you this problem? Or is it from somewhere else? – Toby Mak Oct 07 '19 at 11:00
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https://www.wolframalpha.com/input/?i=sum_%7Bi+%3D+1%7D%5E%7B%5Cinfty%7D+ln%28i%29%2F%28i%28i-ln%28i%29%29%29 – OmG Oct 07 '19 at 11:03
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@Toby Mak, this is from "Problems in Mathematical Analysis 1"by W. J. Kaczor – Mathaddict Oct 07 '19 at 11:19
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To solve the problem, hint is given that $ 0≤\ln n /(n- \ln n) <1, $ so sum is zero. – Mathaddict Oct 07 '19 at 11:23
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I think you mean the sum converges, not that the sum is zero? – Toby Mak Oct 07 '19 at 11:25
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The exact words of book are " observe that $ 0 ≤ \ln n/(n- \ln n) <1 , n \in N $ thus $S = 0$." And they have denoted the sum by $S$ in all other problems. – Mathaddict Oct 07 '19 at 11:29
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So basically you're supposed to check the convergence of the sum? – Zacky Oct 07 '19 at 11:38
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@LeBlanc According to the book, the sum is zero. – Mathaddict Oct 07 '19 at 11:52
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The book is wrong then (see the bellow answer). Or perhaps it was a typo and they wanted to say that the limit of the general term is zero (which is true from what they suggested to observe). – Zacky Oct 07 '19 at 11:53
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Interestingly, numerically this sum is very close to $1+\gamma/2$, where $\gamma$ is the Euler's constant. – Z Ahmed Oct 07 '19 at 12:26
2 Answers
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I checked the actual problem (Problem 3.1.5.(b) of $\text{[1]}$, see also this image). Its exact formulation is as follows:
$$ \sum_{n=1}^{\infty} \frac{1}{n} \left[ \frac{\ln n}{n - \ln n} \right]. $$
Now, in this textbook, the bracket $[\cdot]$ is exclusively used for the greatest integer function. In other words, $[x] = \lfloor x \rfloor$. Now the hint perfectly makes sense, because
$$ 0 \leq \frac{\ln n}{n - \ln n} < 1 \quad \Rightarrow \quad \left\lfloor \frac{\ln n}{n - \ln n} \right\rfloor = 0 $$
and the sum is zero.
$\text{[1]}$ Kaczor, Wiesława J., and Maria T. Nowak. Problems in Mathematical Analysis: Real numbers, sequences, and series. Vol. 1. American Mathematical Soc., 2000., p64
Sangchul Lee
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This should be mentioned with this problem in the book. By the way thanks, now that hint makes sense :). – Mathaddict Oct 07 '19 at 12:02
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$$S=\sum_{n=1}^{\infty} \frac{\ln n}{n(n-\ln n)}= \sum_{n=1}^{\infty} \left( \frac{1} {n-\ln n}- \frac{1}{n} \right)\sim\sum_{n=1}^{\infty} \left( \frac{1}{n} (1+\frac{\ln n}{n}+\frac{\ln^2 n}{n^2}+ \frac{\ln^3 n}{n^3}...) -\frac{1}{n}\right).$$ $$\implies S= \sum_{n=1}^{\infty} \left(\frac{\ln n}{n^2}+\frac{\ln^2 n}{n^3}+ \frac{\ln^3 n}{n^4}+...\right)=-\zeta'(2)+....$$ As per the suggestion of @Toby Mak above.
Further, using $$\sum_{n=0}^{\infty} \frac{1}{n^z}= \zeta(z) \implies \sum_{n=1}^{\infty}(-1)^k \frac{\ln^{k} n}{n^z} =\zeta^{(k)}(z)$$ It comes the same result as predicted by @Sangchui Lee above here in his comment as $$S=\sum_{k=1}^{\infty} (-1)^k \zeta^{(k)} (k+1)$$
However, numeric strongly suggests a value of 1.28495... for S which is very close to $1+\gamma/2~~!$, where $\gamma$ is the Eular number equal to 0.5772...$
Z Ahmed
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I'm amazed the proof is this short, but could you explain your third step? – Toby Mak Oct 07 '19 at 11:21
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The third step is the binomial approxinmation $\frac{1}{n(1-\ln n/n)} \approx \frac{1}{n} (1-\ln n /n)^{-1}.$ – Z Ahmed Oct 07 '19 at 11:39