How Do You Solve Complex Series Without a Calculator?

[anemone]

Here is this week's POTW:

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Without using a calculator, evaluate \(\displaystyle \frac{1}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+\frac{3}{4\times5}+\cdots+\frac{3}{7\times8}+\frac{4}{8\times9}+\cdots+\frac{4}{15\times16}+\cdots+\frac{10}{1023\times1024}.\)

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[MarkFL]

anemone was alerted by an observant member of the community (lfdahl), that there could be a typo in the above problem as originally given. After determining that there was, she asked me to make the correction, as she is at work and wanted this corrected ASAP. I am posting to bring this to the attention of all participants. (Smile)

 

[anemone]

Congratulations to the following members for their correct solution::)

1. castor28
2. kaliprasad
3. lfdahl

Solution from castor28:

Spoiler

The sum (let us call it $S$) consists of blocks of the form:
$$k\left(\frac{1}{2^{k-1}\times(2^{k-1}+1)}+\cdots+\frac{1}{(2^k-1)\times2^k}\right)$$
Because of the identity:
$$\frac{1}{(n-1)n} + \frac{1}{n(n+1)} = \frac{1}{n-1}-\frac{1}{n+1}$$
we have a telescoping sum, and each block is equal to:
$$k\left(\frac{1}{2^{k-1}}-\frac{1}{2^k}\right) = k\,2^{-k}$$
We have therefore
$$S = \sum_{k=1}^{10}k\,2^{-k}$$
If we define
$$f(x) = \sum_{k=1}^{10}x^k2^{-k}$$
we will have:
$$\begin{align*}
f'(x) &= \sum_{k=1}^{10}kx^{k-1}2^{-k}\\
f'(1) &= S
\end{align*}$$
Now, $f(x)$ is a geometric progression with ratio $x/2$. This gives:
$$\begin{align*}
f(x) &= \frac{(x/2)^{11}-(x/2)}{(x/2)-1}\\
&= \frac{x^{11}-1024x}{1024(x-2)}\\
f'(x)&= \frac{1}{1024}\left(\frac{11x^{10}-1024}{x-2} - \frac{x^{11}-1024x}{(x-2)^2}\right)\\
f'(1)&= \frac{1013 + 1023}{1024}\\
S &= \frac{2036}{1024} = {\bf\frac{509}{256}}
\end{align*}$$