0
$\begingroup$

I was trying to prove this equality:

$$ \sum_{i=0}^{n} e^{-\lambda} \frac{\lambda^{i}}{i!}= \frac{1}{n!} \int_{\lambda}^{\infty} e^{-x} x^{n} dx$$

This is my attempt so far:

$$ \frac{1}{n!} \int_{\lambda}^{\infty} e^{-x} x^{n} dx $$ Using integration by parts with $ u = x^{n}, \ du = n \cdot x^{n-1} dx, \ v = -e^{-x}, \ dv = e^{-x} dx $:

$$ \frac{1}{n!} \bigg[ \frac{e^{-x}}{n+1} \ x^{n+1} \bigg\vert_{\lambda}^{\infty} - \int_{\lambda}^{\infty} -n \cdot x^{n-1} e^{-x} dx \bigg]$$

$$ \require{cancel} = \frac{1}{n!} \bigg[\Big( \cancelto{0}{- \frac{x^{n}}{e^{\infty}}} + x^{n} e^{-\lambda}\Big) + n \int_{\lambda}^{\infty} x^{n-1} e^{-x} dx \bigg] $$

Using integration by parts with $ u = x^{n-1}, \ du = (n-1) \cdot x^{n-2} dx, \ v = -e^{-x}, \ dv = e^{-x} dx $: $$ = \frac{1}{n!} \bigg[ x^{n} e^{-\lambda} + n \left(-x^{n-1}e^{-x} \bigg\vert_{\lambda}^{\infty} - \int_{\lambda}^{\infty} - (n-1) \ x^{n-2} dx \right) \bigg] $$ $$ = \frac{1}{n!} \bigg[ x^{n} e^{-\lambda} + n \Big( \cancelto{0}{- \frac{x^{n-1} }{e^{\infty} }} + x^{n-1}e^{-\lambda} + \int_{\lambda}^{\infty} (n-1) \ x^{n-2} dx \Big) \bigg] $$ $$ = \frac{1}{n!} \bigg[ x^{n} e^{\lambda} + nx^{n-1}e^{\lambda} + \int_{\lambda}^{\infty} (n-1) \ x^{n-2} dx \ \bigg] $$

$$ = \frac{1}{n!} \bigg[ \frac{n!}{(n)!} \ x^{n} e^{\lambda} + \frac{n!}{(n-1)!} \ x^{n-1}e^{\lambda} + \int_{\lambda}^{\infty} (n-1) \ x^{n-2} dx \ \bigg] $$

By taking this pattern to its conclusion, we introduce a summation:

$$ = \frac{1}{n!} \bigg[ \sum_{i=0}^{n} \frac{n!}{(n-i)!} x^{n-i} e^{-x} \bigg]_{\lambda}^{\infty} $$ $$ = \bigg[ \sum_{i=0}^{n} \frac{ x^{n-i} \ e^{-x} }{ (n-i)! } \bigg]_{\lambda}^{\infty} $$

At this point I get stuck and don't know how to reach:

$$ \sum_{i=0}^{n} e^{-\lambda} \frac{\lambda^{i}}{i!} $$

I would appreciate it if someone could give me a hint as to where I should be heading now. I can't seem to go any further.

$\endgroup$
2
$\begingroup$

There are some errors in the integrations by parts. We have $$\frac{1}{n!}\int_{\lambda}^{\infty}x^{n}e^{-x}dx=\frac{1}{n!}\left(\left.-e^{-x}x^{n}\right|_{\lambda}^{\infty}+n\int_{\lambda}^{\infty}e^{-x}x^{n-1}dx\right) $$ and so on, then $$\frac{1}{n!}\int_{\lambda}^{\infty}x^{n}e^{-x}dx=\sum_{i=0}^{n}\frac{e^{-\lambda}\lambda^{n-i}}{\left(n-i\right)!} $$ and now use the fact that, for every function $f $, holds $$\sum_{k\leq n}f\left(k\right)=\sum_{k\leq n}f\left(n-k\right). $$

$\endgroup$
1
  • 1
    $\begingroup$ $x$ is the variable of the integral. So you have to consider $x$ at infinity and $x=\lambda$. $\endgroup$ – Marco Cantarini Nov 10 '15 at 6:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.