2
$\begingroup$

Consider $\lambda >0.$ I am reading a paper and the author states that

$$ \displaystyle\lim_{v \rightarrow +\infty} \sum_{n=0}^{+\infty} \frac{\lambda^{n}}{(n !)^v} = 1 + \lambda$$

I tried to compute such limit but I am getting anywhere. Someone could help me?

Thanks in advance!

$\endgroup$
6
$\begingroup$

Note that $$\sum_{n=0}^{+\infty} \frac{\lambda^{n}}{(n !)^v}=1+\lambda+\sum_{n=2}^{+\infty} \frac{\lambda^{n}}{(n !)^v}$$and $$\sum_{n=2}^{+\infty} \frac{\lambda^{n}}{(n !)^v}{\le \sum_{n=2}^{+\infty} \frac{\lambda^{n}}{n !}\sum_{n=2}^{+\infty} \frac{1}{(n !)^{v-1}}\\\le e^{\lambda}\sum_{n=2}^{\infty}{1\over 2^{(n-1)(v-1)}}\\=e^\lambda{{1\over 2^{v-1}}\over 1-{1\over 2^{v-1}}}\\={e^\lambda\over 2^{v-1}-1}\\\to 0}$$hence the result.

$\endgroup$
2
$\begingroup$

The RHS is obviously the first two terms of the sum. For the remaining terms, replace $n!$ by $2^n$. Then, whatever $\lambda$, for a sufficiently large $v$, you have a convergent geometric series (that tends to zero as $v \to \infty$).

$\endgroup$

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.