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The problem:

Given the series $u_n=e^{nx}-e^{(n+1)x}$ find $\sum_{n=1}^N u_n$ in terms of n and x.

Find the set of values which the infinite series converges, and give the sum to infinity.

I did the limit test, and I am stuck at how to find the limit at x=0.

Any suggestions?

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1 Answer

up vote 3 down vote accepted

This sum telescopes, giving partial sums $$s_N=\sum_{n=1}^N u_n=(e^x-e^{2x})+(e^{2x}-e^{3x})+\ldots+(e^{Nx}-e^{(N+1)x})=e^x-e^{(N+1)x}.$$ If $x<0$, then $s_N\to e^x$ as $N\to\infty$.

If $x=0$, then $s_N=0$ identically for all $N$, so $s_N\to 0$.

If $x>0$, then the sum diverges.

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so why don't you say it converges for $ x \le 0$ ? Doesn't the properties of $e^x$ contain 1 & 2? Why do you have a separate case for $x = 0$? –  yiyi Sep 19 '12 at 0:40
    
Because $e^0=1$ –  Jonathan Sep 19 '12 at 5:47
    
so it converges when x=0, just making sure I fully understand the usages of the word convergence. –  yiyi Sep 20 '12 at 0:32
    
That's correct, the series converges for $x\le 0$. –  Jonathan Sep 20 '12 at 0:49
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