1
$\begingroup$

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?

1 Answers 1

3

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.

  • 0
    That's correct, the series converges for $x\le 0$.2012-09-20