Is there a simpler expression for the sum $\sum_{i=1}^{\infty} \frac{H_{i}}{i+1}$?

Question

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!

Answer 1

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 $\sim\int_1^n\frac{\ln xdx}{x}=\frac12\ln^2n$.

Answer 2

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.

Answer 3

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 : $$\sum_{i=1}^\infty \frac{H_i}{i+1}=\int_0^1\frac{\ln(1-t)}{t-1}dt$$