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How do I calculate the sum of $$\sum_{n=0}^{\infty}{x^{n+2}\over (n+2)n!}=\sum_{n=0}^{\infty}\color{red}{x^{n}\over n!}\cdot{{x^2\over n+2}}?\tag1$$

$$\sum_{n=0}^{\infty}{x^{n}\over n!}=e^x\tag2$$

José Carlos Santos
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3 Answers3

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If $f(x)=\sum_{n=0}^\infty\frac{x^{n+2}}{(n+2)n!}$, then $f'(x)=\sum_{n=0}^\infty\frac{x^{n+1}}{n!}=xe^x$. Therefore, since $f(0)=0$, $f(x)=(x-1)e^x+1$.

José Carlos Santos
  • 427,504
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Hint:

$$\dfrac{x^{n+2}}{(n+2) n!}=\dfrac{(n+1)x^{n+2}}{(n+2)!}=\dfrac{(n+2-1)x^{n+2}}{(n+2)!}$$

$$=x\cdot\dfrac{x^{n+1}}{(n+1)!}-\dfrac{x^{n+2}}{(n+2)!}$$

$$\implies\sum_{n=0}^\infty\dfrac{x^{n+2}}{(n+2) n!}=x\sum_{n=0}^\infty\dfrac{x^{n+1}}{(n+1)!}-\sum_{n=0}^\infty\dfrac{x^{n+2}}{(n+2)!}$$

Now use $e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$

lab bhattacharjee
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  • $\sum_{r=0}^{\infty}{x^{r+1}\over (r+1)!}=?$ @Lab... –  Apr 03 '18 at 06:55
  • @GiantLeap, $$\sum_{n=0}^\infty\dfrac{x^{n+1}}{(n+1)!}=-\dfrac{x^{-1+1}}{(-1+1)!}+\sum_{n=-1}^\infty\dfrac{x^{n+1}}{(n+1)!}=?$$ – lab bhattacharjee Apr 03 '18 at 06:56
  • (+1) Nice answer! And better than mine. – José Carlos Santos Apr 03 '18 at 09:25
  • See also : https://math.stackexchange.com/questions/581603/evaluate-frac13-frac14-frac12-frac15-frac13-dots and https://math.stackexchange.com/questions/2638073/how-to-solve-this-summation-without-taylor and https://math.stackexchange.com/questions/576976/evaluate-the-series-lim-limits-n-to-infty-sum-limits-i-1n-fracn22 and https://math.stackexchange.com/questions/845464/determine-the-value-of-the-series-sum-n-1-infty-frac1-n2-n – lab bhattacharjee Apr 03 '18 at 09:31
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$$\sum_{n\geq 0}\frac{x^{n+2}}{(n+2)n!}=\sum_{n\geq 0}\int_{0}^{x}\frac{z^{n+1}}{n!}\,dz = \int_{0}^{x}\sum_{n\geq 0}\frac{z^{n+1}}{n!}\,dz = \int_{0}^{x} z e^z\,dz=\left[(z-1)e^z\right]_{0}^{x}. $$ The exchange of $\sum$ and $\int$ is allowed by absolute convergence.

Jack D'Aurizio
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