Question is to Evaluate : $$\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots$$

what all i could do is :

$$\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots=\sum_{n=1}^{\infty} \frac{1}{(n+2)n!}=\sum_{n=1}^{\infty} \frac{n+1}{(n+2)!}=\sum_{n=1}^{\infty} \frac{n}{(n+2)!}+\sum_{n=1}^{\infty} \frac{1}{(n+2)!}$$

I have $$\sum_{n=1}^{\infty} \frac{1}{(n+2)!}=\sum_{n=0}^{\infty} \frac{1}{n!}-1-\frac{1}{2}=e-\frac{3}{2}$$

Now, I am not able to see what $$\sum_{n=1}^{\infty} \frac{n}{(n+2)!}$$ would be.

I would be thankful if some one can help me to clear this.

Thank you.


$$\text{From }\frac{n+1}{(n+2)!},$$ $$=\frac{n+2-1}{(n+2)!}=\frac1{(n+1)!}-\frac1{(n+2)!}$$

Can you identify the Telescoping series?

The survivor will be $$\frac1{2!}$$

  • 2
    $\begingroup$ @PraphullaKoushik, The other methods are unnecessarily complex unless I am missing something $\endgroup$ – lab bhattacharjee Nov 26 '13 at 8:37
  • $\begingroup$ Perfect i would say.. :) :) Thank you.. :) $\endgroup$ – user87543 Nov 26 '13 at 8:48
  • $\begingroup$ @PraphullaKoushik, sometimes, formulae misdirect us to complexity $\endgroup$ – lab bhattacharjee Nov 26 '13 at 8:50
  • $\begingroup$ yes yes.. totally agree :D $\endgroup$ – user87543 Nov 27 '13 at 4:40

Hint : $\displaystyle f(x)=\sum\frac{x^n}{(n+1)!}\iff f'(x)=\sum\frac{n\cdot x^{n-1}}{(n+1)!}\quad,\quad S=f'(1)$.

  • $\begingroup$ but i need $n+2$ in denominator.. :( do i have to take $f(x)=\sum \frac{x^n}{(n+2)!}$?? This does help but i do not know what is compact form for $\sum \frac{x^n}{(n+2)!}$ I only know that $e^x=\sum \frac{x^n}{(n)!}$ $\endgroup$ – user87543 Nov 26 '13 at 7:05
  • $\begingroup$ Start at $n=2$. $\endgroup$ – Lucian Nov 26 '13 at 7:06
  • $\begingroup$ I am sorry i could not make this out from hint :( $\endgroup$ – user87543 Nov 26 '13 at 7:12
  • $\begingroup$ @PraphullaKoushik: $\dfrac n{(n+1)!}=\dfrac n{(n-1)!~n(n+1)}=\dfrac1{(n-1)!~(n+1)}$ $\endgroup$ – Lucian Mar 10 '17 at 20:42

Since $\displaystyle e^x=\sum_{n=0}^\infty \frac {x^n}{n!}$, then $\displaystyle e^x-1-x=\sum_{n=0}^\infty\frac{x^{n+2}}{(n+2)!}$, and $$ \frac{e^x-1-x}{x}=\sum_{n=0}^\infty\frac{x^{n+1}}{(n+2)!}, $$ so $$ \left(\frac{e^x-1-x}{x}\right)'=\sum_{n=0}^\infty\frac{(n+1)x^n}{(n+2)!}. $$ Evaluate at $x=1$, and subtract $1/2$ (the term corresponding to $n=0$) to find the value you are after. You obtain $$ \frac12=1-\frac12=\left(\frac{e^x(x-1)+1}{x^2}\right)_{x=1}-\frac12=\sum_{n=1}^\infty\frac{n+1}{(n+2)!}. $$

(By the way, you have a small typo in your analysis, since $\displaystyle e=\sum_{n=0}^\infty\frac1{n!}$; on line $6$ you wrote the series starting at $n=1$.)

  • $\begingroup$ oh.. This is so helpful... Thank you.... $\endgroup$ – user87543 Nov 26 '13 at 7:09
  • $\begingroup$ You are welcome. Glad I could help. $\endgroup$ – Andrés E. Caicedo Nov 26 '13 at 7:10
  • $\begingroup$ Thanks for mentioning the typo.. It is corrected now :) $\endgroup$ – user87543 Nov 26 '13 at 7:11



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