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.

How to find the following series' value?

$$\sum_{n=1}^{\infty}\bigg(e-\Big(1+\frac{1}{n}\Big)^n\bigg)$$

share|improve this question
    
how do you know if the series converges? –  Aang Mar 3 '13 at 22:52
5  
It converges because my book makes me find the value. –  Guillermo Mar 3 '13 at 22:56
6  
Dear Ryuichi, If your book told you to jump off a bridge, would you do it? ;-) –  Bruno Joyal Mar 3 '13 at 22:57
    
Mathematica tells me the sum does not converge. I don't know how realiable this is but at least we know it's not some standard sum –  muzzlator Mar 3 '13 at 23:06

3 Answers 3

up vote 5 down vote accepted

The telescoping series $$e-\left({1+{1\over n}}\right)^n=\sum_{j=1}^n\left({1+{1\over n}}\right)^{j-1}\left[\exp(1/n)-\left({1+{1\over n}}\right)\right] \exp((n-j)/n)$$ shows that $$e-\left({1+{1\over n}}\right)^n\geq n \left[\exp(1/n)-\left({1+{1\over n}}\right)\right]\geq n \,{1\over 2}\left({1\over n}\right)^2 = {1\over 2n}$$ for all $n\geq 1$. Therefore the OP's series diverges by comparison with the harmonic series.

share|improve this answer

The sum is diverging as for $n>2$ $$e-\left(1+\frac{1}{n}\right)^n > \frac{1}{n}$$

share|improve this answer
2  
Well, for $n>2$ anyway.... –  Byron Schmuland Mar 3 '13 at 23:12
    
oh thanks, forgot the $n>2$ –  Dominic Michaelis Mar 3 '13 at 23:13

If the book says it converges, how about the value of

$$\sum_{n=1}^{\infty}\bigg(e-\Big(1+\frac{1}{n}\Big)^{n+1/2}\bigg)$$

(which does converge)?

share|improve this answer

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.