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.

$$ u_{n}=-n+\sum_{k=1}^n e^{\frac{1}{k+n}}$$

$$ u_{n+1}-u_{n}=-1+\exp\left(\frac{1}{2n+2}\right)+\exp\left(\frac{1}{2n+1}\right)-\exp\left(\frac{1}{n+1}\right)=O(1/n^3)$$

So $ \sum u_{n+1}-u_n$ and $u_n$ converge.

$ u_{100000} \approx 0.69 $

The limit seems to be $\ln2$

share|improve this question
    
Is it $\left[\sum_{k=1}^n e^{\frac{1}{k+n}}\right]-n$ or $\sum_{k=1}^n \left[e^{\frac{1}{k+n}}-n\right]$? –  akkkk May 28 '12 at 15:44
    
Your formula for $u_{n+1}-u_n$ is wrong. –  Did May 28 '12 at 15:49
    
Obviously $u_{n}=-n+\sum_{k=1}^n e^{\frac{1}{k+n}}$ –  Chon May 28 '12 at 15:52
    
I have corrected $u_{n+1}-u_{n}$; the series is convergent... –  Chon May 28 '12 at 16:07
    
Two people have answered, and so far the only up-vote is from me. –  Michael Hardy May 28 '12 at 16:11

2 Answers 2

up vote 5 down vote accepted

For every $x\leqslant1$, $1+x\leqslant\mathrm e^x\leqslant1+x+x^2$ hence $u_n=v_n+w_n$ with $$ v_n=\sum_{k=1}^n\frac1{k+n}=\frac1n\sum_{k=1}^n\frac1{\frac{k}n+1},\qquad 0\leqslant w_n\leqslant\sum_{k=1}^n\frac1{(k+n)^2}\leqslant\frac1n. $$ In particular, $w_n\to0$. On the other hand, $v_n$ is the $n$th Riemann sum of the function $x\mapsto\frac1{1+x}$ on the interval $[0,1]$ hence $v_n\to\int\limits_0^1\frac{\mathrm dx}{1+x}=\log2$. Finally, $\lim\limits_{n\to\infty}u_n=\log2$.

share|improve this answer
1  
You can also use $e^x \leq \tfrac{1}{1-x}$ for $x < 1$. This leads to $$ u_n \leq \sum_{k=1}^n \frac{1}{k + n - 1}. $$ –  WimC May 28 '12 at 17:19

Everything boils down to the n times use of the following elementary limit, namely: $$\lim_{x\to0}\frac{e^{x}-1}{x}=1$$

Consequently, by expanding the sum we have that:

$$\lim_{n\to\infty}\frac{e^\frac{1}{n+1}-1}{\frac{1}{n+1}} \frac{1}{n+1} + \lim_{n\to\infty}\frac{e^\frac{1}{n+2}-1}{\frac{1}{n+2}} \frac{1}{n+2}+ \cdots = \lim_{n\to\infty}\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} = $$ $$\lim_{n\to\infty}{\gamma}+\ln{2n}-{\gamma}-\ln{n}= \ln{2}.$$ Also notice that i've just applied another well-known limit: $$\lim_{n\to\infty} 1+\frac1{2}+\cdots+\frac{1}{n}-\ln{n}={\gamma}$$ $\tag{$\gamma$ is Euler-Mascheroni constant}$

The proof is complete.

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.