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Erfan's question at Yahoo! Answers regarding summation of series

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  • #1


Staff member
Feb 24, 2012
Here is the question:

Summation of series ( Method of differences )?

Show that r/(r+1)! = 1/r! - 1/(r+1)! , hence or otherwise , evaluate i) sum of r/(r+1)! from 1 to n

ii) sum of (r+2)/(r+1)! from 1 to infinity
giving your answer to part ii in the terms of e .
I have posted a link there to this topic so the OP can see my work.
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Feb 24, 2012
Hello again Erfan,

First we are asked to show:

\(\displaystyle \frac{r}{(r+1)!}=\frac{1}{r!}-\frac{1}{(r+1)!}\)

There are a couple of ways we could do this:

a) Combine terms on the right side:

\(\displaystyle \frac{1}{r!}-\frac{1}{(r+1)!}=\frac{(r+1)!-r!}{r!(r+1)!}=\frac{r!((r+1)-1)}{r!(r+1)!}=\frac{r}{(r+1)!}\)

b) Add \(\displaystyle 0=1-1\) to the numerator on the left side:

\(\displaystyle \frac{r}{(r+1)!}=\frac{(r+1)-1}{(r+1)!}=\frac{1}{r!}-\frac{1}{(r+1)!}\)

i) We are asked to evaluate:

\(\displaystyle S_n=\sum_{r=1}^n\left(\frac{r}{(r+1)!} \right)\)

Using the result above, we may write:

\(\displaystyle S_n=\sum_{r=1}^n\left(\frac{1}{r!} \right)-\sum_{r=1}^n\left(\frac{1}{(r+1)!} \right)\)

Re-indexing the first sum, we have:

\(\displaystyle S_n=\sum_{r=0}^{n-1}\left(\frac{1}{(r+1)!} \right)-\sum_{r=1}^n\left(\frac{1}{(r+1)!} \right)\)

Pulling off the first term from the first sum and the last term from the second, we will be left with sums having the same indices:

\(\displaystyle S_n=\left(1+\sum_{r=1}^{n-1}\left(\frac{1}{(r+1)!} \right) \right)-\left(\sum_{r=1}^{n-1}\left(\frac{1}{(r+1)!} \right)+\frac{1}{(n+1)!} \right)\)

The sums add to zero, and we are left with:

\(\displaystyle S_n=1-\frac{1}{(n+1)!}\)

ii) We are now asked to evaluate:

\(\displaystyle S_{\infty}=\sum_{r=1}^{\infty}\left(\frac{r+2}{(r+1)!} \right)\)

We may rewrite the summand to obtain:

\(\displaystyle S_{\infty}=\sum_{r=1}^{\infty}\left(\frac{r}{(r+1)!} \right)+2\sum_{r=1}^{\infty}\left(\frac{1}{(r+1)!} \right)\)

Using the result from part i) for the first sum and re-indexing the second sum, there results:

\(\displaystyle S_{\infty}=\lim_{n\to\infty}\left(1-\frac{1}{(n+1)!} \right)+2\left(\sum_{r=2}^{\infty}\left(\frac{1}{r!} \right) \right)\)

Evaluating the limit and rewriting the sum, we find:

\(\displaystyle S_{\infty}=1+2\left(\sum_{r=0}^{\infty}\left(\frac{1}{r!} \right)-2 \right)\)


\(\displaystyle e=\sum_{r=0}^{\infty}\left(\frac{1}{r!} \right)\)

we now have:

\(\displaystyle S_{\infty}=1+2\left(e-2 \right)=2e-3\)