Which test?????? I dunno. Convergence is quite tricky.

FranklinMint · **Joined:** Thu, 7 Aug 2008 12:59:30 UTC **Posts:** 15

Which convergence test would you use to determine whether this series, $$\sum_{n=1}^{\infty}\text{\huge (}\frac{2}{3}+e^{-n}\text{\huge)}^n$ converges.

How do you know which one to use. And could you show me how this particlular example works please???????!!!!!!!!!

Sometimes maths is a pain.

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9632

For n > 10, we have e^(-n) < 1/6.

Thus $\left(\frac23+e^{-n}\right)^n<\left(\frac56\right)^n$ for sufficiently large n.

FranklinMint · **Joined:** Thu, 7 Aug 2008 12:59:30 UTC **Posts:** 15

Matt wrote:

For n > 10, we have e^(-n) < 1/6.

Thus $\left(\frac23+e^{-n}\right)^n<\left(\frac56\right)^n$ for sufficiently large n.

Is that using any test im not aware of, or is that just by deduction?

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9632

It's the comparison test (also known as direct comparison test).

S.O.S. Mathematics CyberBoard

Which test?????? I dunno. Convergence is quite tricky.

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