Summation of a series-combination of a GP and another series

Question

Find the sum of the series:

$$\frac{1}{2}\biggl(\frac{1}{2}\biggl)+\frac{2}{3}\biggl({\frac{1}{2}\biggl)}^2+\frac{3}{4}\biggl({\frac{1}{2}\biggl)}^3+\cdots+ n^{th}terms$$

I calculated the $n^{th}$ term to be $\frac{n}{n+1}\frac{1}{2^n}$ $$\sum_{k=1}^n \frac{k}{k+1}\frac{1}{2^k}=\sum_{k=1}^n \frac{1}{2^k}\biggl(1-\frac{1}{k+1}\biggl)=\sum_{k=1}^n \frac{1}{2^k}-\sum_{k=1}^n \frac{1}{2^k(k+1)} $$

The first summation is a GP. But then I could not solve the second summation as it was not telescoping. Please help.

@See-Woo Lee No never heard of it. Can u specify some good site for it?? — user452508, Sep 27 '17 at 07:42
Is it enough for you to find $\sum_{k=1}^{\infty}t_k$ (then e.g. have a look here) or do you also need an expression for $\sum_{k=1}^{n}t_k$? — drhab, Sep 27 '17 at 07:49

score 1 · Answer 1 · answered Sep 27 '17 at 08:08

1

You have $\frac{1}{1-x} = \sum_{i=0}^\infty x^i$

So $-\ln(1-x) = \sum_{i=0}^\infty \frac{1}{i+1}x^{i+1}$

Then $$\frac{-\ln(1-x)}{x} = \sum_{i=0}^\infty \frac{1}{i+1}x^{i}$$

So, you have $$\frac{-\ln(\frac{1}{2})}{\frac{1}{2}} = 1 + \sum_{i=1}^\infty \frac{1}{i+1}\frac{1}{2^i}$$

answered Sep 27 '17 at 08:08

GAVD

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The upper limit is $n$, not $\infty$. – B. Goddard Sep 27 '17 at 12:50
From where does log come?? – user452508 Sep 27 '17 at 13:21

score 1 · Answer 2 · answered Sep 27 '17 at 09:09

If we consider the more general case of $$S_n(x)=\sum_{k=1}^n \frac{k}{k+1}{x^k}=\sum_{k=1}^n x^k\biggl(1-\frac{1}{k+1}\biggl)=\sum_{k=1}^n x^k-\sum_{k=1}^n \frac{x^k}{(k+1)}$$ the problem looks quite simple when the summation is done up to $\infty$.

For finite $n$, at least to me, the problem seems to be difficult since we have $$\sum_{k=1}^n x^k=\frac{x \left(1-x^n\right)}{1-x}$$ $$\sum_{k=1}^n \frac{x^k}{(k+1)}=-1-\frac{\log (1-x)}{x}-x^{n+1} \Phi (x,1,n+2)$$ where appears the Lerch transcendent function. This makes

$$S_n(x)=1+\frac{x \left(1-x^n\right)}{1-x}+\frac{\log (1-x)}{x}+x^{n+1} \Phi (x,1,n+2)$$ $$S_n\left(\frac 12\right)=2-2\log(2)+\frac{\Phi \left(\frac 12,1,n+2\right)}{2^{n+1}}$$ As shown below, the last term decreases very fast $$\left( \begin{array}{cc} n & \frac{\Phi \left(\frac 12,1,n+2\right)}{2^{n+1}}\\ 1 & 0.136294 \\ 2 & 0.052961 \\ 3 & 0.021711 \\ 4 & 0.00921103 \\ 5 & 0.00400269 \\ 6 & 0.00177055 \\ 7 & 0.000793989 \\ 8 & 0.000359961 \\ 9 & 0.000164649 \\ 10 & 0.0000758704 \end{array} \right)$$ and a quick and dirty nonlinear regression (done for $1\leq n \leq 20$) seems to show a quite good approximation by $$\frac{\Phi \left(\frac 12,1,n+2\right)}{2^{n+1}}\approx 0.414102\, e^{-1.11129\, n^{0.888026}}$$

How did you take such a sudden log – user452508 Sep 27 '17 at 13:20 — user452508, Sep 27 '17 at 13:20

Summation of a series-combination of a GP and another series

2 Answers2