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I've been dealing probability questions lately, and I had to calculate this sum : $\sum_{k=0}^\infty e^{-3} \frac{k}{k+1} \frac{3^k}{k!} $ , which I have no idea how to bring to to an easier form, or for example extract from it something that is known. Its good to mention that its a result of multiplying with Poisson function of probability with parameter $3$.

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    $\begingroup$ $\frac{1}{k!}\frac{k}{k+1} = \frac{1}{k!}-\frac{1}{(k+1)!}$ $\endgroup$ – reuns Feb 26 '17 at 0:50
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Note that if we let $k=n-1$,

$$\frac k{k+1}\frac{3^k}{k!}=\frac{k3^k}{(k+1)!}=\frac13\frac{(n-1)3^n}{n!}=\frac13\left(\frac{3^n}{(n-1)!}-\frac{3^n}{n!}\right)$$

Summing from $n=1$ to infinity yields

$$3e^3S=\sum_{n=1}^\infty\left(\frac{3^n}{(n-1)!}-\frac{3^n}{n!}\right)=3e^3-e^3+1=1+2e^3$$

Thus,

$$\sum_{k=0}^\infty e^{-3}\frac k{k+1}\frac{3^k}{k!}=\frac{1+2e^3}{3e^3}$$

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  • $\begingroup$ I'm so sorry. i had edited the question. $\endgroup$ – Firas Ali Abdel Ghani Feb 25 '17 at 22:02
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    $\begingroup$ Beautifully answered! $\endgroup$ – Sijaan Hallak Feb 25 '17 at 22:16
  • $\begingroup$ You really treat this as Art! $\endgroup$ – Firas Ali Abdel Ghani Feb 25 '17 at 22:21
  • $\begingroup$ :-) Thanks guys. $\endgroup$ – Simply Beautiful Art Feb 25 '17 at 22:36

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