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I'm trying to find a formula for the partial sum of $n/2^n$.

I've tried this so far...

$$ S_n = \frac12 + \frac24 + \frac38 + \cdots + \frac{n}{2^n} $$

Then I tried to find a way to eliminate most of the terms by multiplying the whole sequence by $\frac{2^n}n\cdot\frac{n+1}{2^{n+1}}$ (a terms that should take each term in the sequence to the next term.)

$$ \frac{n+1}{2n} S_n = \frac24 + \frac38 + \cdots + \frac{n+1}{2^{n+1}} $$

Then

$$ S_n- \frac{n+1}{2n} S_n = \frac12 - \frac{n+1}{2^{n+1}}$$

$$ S_n = \frac{\frac12 - \frac{n+1}{2^{n+1}}}{1-\frac{n+1}{2n}} $$

But, alas, this does not give the correct answers.

Can someone point out where I went wrong? Thanks

Ryan Stull
  • 489

3 Answers3

5

Just multiply by $\frac12$, not that more complicated factor. $$\frac12S_n=\frac14+\frac28+\frac3{16}+\dots+\frac n{2^{n+1}}$$ $$S_n-\frac12S_n=\frac12+\frac14+\frac18+\dots+\frac1{2^n}-\frac n{2^{n+1}}$$ $$\frac12S_n=1-\frac1{2^n}-\frac n{2^{n+1}}$$ $$S_n=2\left(1-\frac1{2^n}-\frac n{2^{n+1}}\right)$$

Parcly Taxel
  • 103,344
2

Observe

$$2S_n = 1+ \frac{1+1}2 + \frac{2+1}4 + \cdots + \frac{(n-1)+1}{2^{n-1}} \\=S_{n-1}+1+\frac12+\frac14\cdots+\frac1{2^{n-1}} \\=S_n-\frac n{2^n}+2-\frac1{2^{n-1}}.$$

1

Hint

Consider $$S_n=\sum_{k=1}^n k x^k=x\sum_{k=1}^n k x^{k-1}=x \left(\sum_{k=1}^n x^{k} \right)'$$

Just finish and make $x=\frac 12$.

Claude Leibovici
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