1. Find a formula for the sum

2. Find the sum of

3. Find a formula for

Can anyone please solve step by step, I just started summations and need to see the procedure of doing these questions

Thanks!

2. For #1: It can be rewritten as $\displaystyle \sum_{k=1}^{\infty}\frac{1}{k!}-\sum_{k=1}^{\infty}\frac{1}{(k+1)!}$

The first one, the sum of the reciprocals of the factorials, is a well-known sum which equals e-1 and can be derived from the Taylor series for e by letting x=1.

The second one $\displaystyle \sum_{k=1}^{\infty}\frac{1}{(k+1)!}=e-2$

$\displaystyle e-1-(e-2)=1$

3. Originally Posted by usagi_killer
1. Find a formula for the sum

2. Find the sum of

$\displaystyle \log_ku=\frac{\ln u}{\ln k}\Longrightarrow \sum\limits_{k=2}^n\frac{1}{\log_ku}=$ $\displaystyle \frac{1}{\ln u}\sum\limits_{k=2}^n\ln k=\frac{1}{\ln u}\ln\left(\prod\limits_{k=2}^nk\right)=\frac{\ln k!}{\ln u}=\log_uk!$

Tonio

3. Find a formula for

Can anyone please solve step by step, I just started summations and need to see the procedure of doing these questions

Thanks!
.

4. Originally Posted by usagi_killer
3. Find a formula for
$\displaystyle \sum\limits_{j=1}^{n}{\frac{1}{j(j+1)(j+2)}}=\frac {1}{2}\sum\limits_{j=1}^{n}{\frac{(j+2)-j}{j(j+1)(j+2)}},$ hence your sum equals $\displaystyle \frac{1}{2}\sum\limits_{j=1}^{n}{\left( \frac{1}{j(j+1)}-\frac{1}{(j+1)(j+2)} \right)}=\frac{1}{2}\left( \frac{1}{2}-\frac{1}{(n+1)(n+2)} \right).$

5. Originally Posted by galactus
For #1: It can be rewritten as $\displaystyle \sum_{k=1}^{\infty}\frac{1}{k!}-\sum_{k=1}^{\infty}\frac{1}{(k+1)!}$

The first one, the sum of the reciprocals of the factorials, is a well-known sum which equals e-1 and can be derived from the Taylor series for e by letting x=1.

The second one $\displaystyle \sum_{k=1}^{\infty}\frac{1}{(k+1)!}=e-2$

$\displaystyle e-1-(e-2)=1$
Thanks for the help! Is there a way to do this without Taylor's series and how did you get the 2nd summation?

6. Originally Posted by usagi_killer
Thanks for the help! Is there a way to do this without Taylor's series and how did you get the 2nd summation?

Galactus did a nice job...but with an infinite series, whereas you wanted a finite one, yet the answer STILL is in Galactus message:

$\displaystyle \sum\limits_{k=1}^n\frac{k}{(k+1)!}=\sum\limits_{k =1}^n\frac{1}{k!}-\sum\limits_{k=1}^n\frac{1}{(k+1)!}=1-\frac{1}{(n+1)!}=\frac{(n+1)!-1}{(n+1)!}$

Tonio
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