Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Test the convergence of the series
$$\sum_{n=1}^{\infty}\frac{(1+\frac{n}{\mathrm{e}})^n}{n\mathrm{!}}$$
Thanks!

share|improve this question
    
What tests have you tried? What can you say about the numerator? –  Calvin Lin Jan 29 '13 at 7:53
    
I have tried cauchy root test, it gives no conclusion. –  rajkamalmath Jan 29 '13 at 7:55
2  
You will find the Stirling approximation to the factorial works very nicely. One could get away with somewhat less. –  André Nicolas Jan 29 '13 at 7:58
add comment

2 Answers

up vote 12 down vote accepted

Observe that by Stirling's approximation we always have $$n!\le e\sqrt{n}\left(\frac{n}{e}\right)^n$$ and so we see that $$\sum_{n=1}^\infty\frac{\left(1+\frac{n}{e}\right)^n}{n!}\ge \sum_{n=1}^\infty\frac{\left(\frac{n}{e}\right)^n}{e\sqrt{n}\left(\frac{n}{e}\right)^n}=\frac{1}{e}\sum_{n=1}^\infty\frac{1}{\sqrt{n}}=\infty$$ thus the series diverges.

share|improve this answer
add comment

we have $$\left(\frac{n}{\mathrm{e}}\right)\leq \left(1+\frac{n}{\mathrm{e}}\right)$$
And so we have$$\frac{{\left(\frac{n}{\mathrm{e}}\right)}^n}{n\mathrm !}\leq \frac{{\left(1+\frac{n}{\mathrm{e}}\right)}^n}{n\mathrm !}$$
Now the series $\sum_{n=1}^{\infty}\frac{{\left(\frac{n}{\mathrm{e}}\right)}^n}{n\mathrm !}$ is divergent by logarithmic test. So the given series is divergent.

share|improve this answer
    
I think doing in this way would be much easier. –  rajkamalmath Jan 29 '13 at 8:20
3  
Why are you asking this question if you already have a solution? -2. –  Benjamin Dickman Jan 29 '13 at 8:22
    
I have found it by seeing the previous answer. To be more precise by seeing 2nd step of it. –  rajkamalmath Jan 29 '13 at 8:23
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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