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Erfan's question at Yahoo! Answers regarding summation of series
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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:
\(\displaystyle S_{\infty}=1+2\left(\sum_{r=0}^{\infty}\left(\frac{1}{r!} \right)-2 \right)\)
Using:
\(\displaystyle e=\sum_{r=0}^{\infty}\left(\frac{1}{r!} \right)\)
we now have:
\(\displaystyle S_{\infty}=1+2\left(e-2 \right)=2e-3\)
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:
\(\displaystyle S_{\infty}=1+2\left(\sum_{r=0}^{\infty}\left(\frac{1}{r!} \right)-2 \right)\)
Using:
\(\displaystyle e=\sum_{r=0}^{\infty}\left(\frac{1}{r!} \right)\)
we now have:
\(\displaystyle S_{\infty}=1+2\left(e-2 \right)=2e-3\)