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Is there any way to simplify the following series?

$$\frac{1}{2}(1) + \frac{1}{3}(1 + \frac{1}{2}) + \frac{1}{4}(1+\frac{1}{2}+\frac{1}{3}) + \frac{1}{5}(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}) + \dots$$

This could be written

$$ \sum_{i=1}^{\infty} \frac{H_{i}}{i+1} $$

where $H_i=\sum_{k=1}^i\frac1k$ is the ith harmonic number.

I'm wondering if there's a simpler way to write this, so there is only one summation and not two?

EDIT: everyone is pointing out it diverges. I'm just wondering if it can be written more simply. For example, the sum or product of harmonic numbers or something like that.

EDIT 2: lots of great answers and suggestions so far. Jean-Claude Arbaut suggested plugging the above into Wolfram; what comes out is $\sum_{i=1}^\infty = H_i / (i+1) = \frac{1}{2}(H_{n+1}^2 + H_{n+1}^{(2)})$ where $H^{(r)}$ is a generalized harmonic number. I'd be very interested to get some pointers on how to do that by hand? Also, I notice that expression looks an awful lot like the $\frac{1}{2} \ln^2{n}$ of J.G's answer.

BTW the context for all this is is that this particular expression came out the solution to a recursion; I have a whole set of related recursions and I'm trying to understand the overall pattern to their solutions. Thanks everybody!

Ben S.
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    Doesn't converge. – Jean-Claude Arbaut May 11 '20 at 15:56
  • Since $H_i \sim \log i$, the series diverges. – Gary May 11 '20 at 15:58
  • I guess I was just wondering, even if it diverges, is there a way to write it with just one summation (since the H_i is itself a summation)? Does it simplify? Or to put it differently, can it somehow cleverly be written only in terms of harmonic numbers without any other summations? – Ben S. May 11 '20 at 16:01
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    If that's what you want: https://www.wolframalpha.com/input/?i=Sum%5BHarmonicNumber%5Bk%5D%2F%28k%2B1%29%2C%7Bk%2C1%2Cn%7D%5D – Jean-Claude Arbaut May 11 '20 at 16:01
  • @Ben S. Please first show us your own effort, possible intermediate results and where you are stuck.You have already stated an interesting conjecture in your comment. Following this idea should bring you to the solution by yourself. Good luck. – Dr. Wolfgang Hintze May 12 '20 at 11:12

3 Answers3

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To expand on @Jean-ClaudeArbaut's point, since $H_i$ is at least $O(1)$ (indeed it's $\sim\ln i$), a comparison with the harmonic series implies the given series diverges. In fact, the $n$th partial sum is $\sim\int_1^n\frac{\ln xdx}{x}=\frac12\ln^2n$.

J.G.
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The series $\sum_{i=1}^\infty \frac{1}{i+1}$ does not converge; it is the harmonic series (without its first term). Your series has each term larger than $\frac{1}{i+1}$, so by direct comparison, your series diverges.

Eric Towers
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you can use the generation function of harmonic numbers : $$\sum_{i=1}^\infty H_iz^i=\frac{\ln(1-z)}{z-1}$$ Then you have : $$\sum_{i=1}^\infty \frac{H_i}{i+1}=\int_0^1\frac{\ln(1-t)}{t-1}dt$$

Achraf BOURASS
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