5

Find out the sum of the following infinite series $$\frac{3}{2^2(1)(2)} + \frac{4}{2^3(2)(3)} +\dots+\frac{r+2}{2^{r+1}(r)(r+1)}+\cdots $$ up to $r\to\infty$.

MY TRY:- I tried to split $r+2$ as $[(r+1) +{(r+1)-r}]$ so that I can cancel one term from each terms in the numerator. Then I got an expression which was like Harmonic-Geometric series. But I could not do further any more after this.

AccidentalFourierTransform
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Priyanka Vithani
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  • Hint: rth term is same as $\frac{1}{r 2^r}-\frac{1}{ (r+1) 2^{r+1}}$ so you were already pretty close. A split of $2*(r+1)-r$ thus would have been more beneficial. – Someone Jul 13 '17 at 11:49

2 Answers2

9

If you do the partial fraction expansion, the summand becomes

$$\frac{1}{2^{r+1}}\left( \frac{2}{r} - \frac{1}{r+1}\right) = \frac{1}{2^r r} - \frac{1}{2^{r+1}(r+1)}, $$

So the sequence is telescoping. All terms cancel except the first and so sum equals $\frac12$.

Astyx
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B. Goddard
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3

prove by induction that for your sum is hold $$\sum_{i=1}^n\frac{i+2}{2^{i+1}i(i+1)}=\frac{2^{-n-1} \left(2^n n+2^n-1\right)}{n+1}$$

Dr. Sonnhard Graubner
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