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So the question is how can I see if this series :

$$\displaystyle\sum_{n=1}^{\infty} \frac{1}{(4-(-1)^n)^n}$$

converges and find its sum.

So I would probably need to use the Leibnitz criterion for alternating series, but I'm not 100% if this series is an alternating one. Any help would be appreciated.

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4 Answers 4

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$$ \frac{1}{(4-(-1)^n)^n}\leq \frac{1}{3^n}$$

And comparison test.

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It's a positive series, clearly not an alternating one!

Consider the terms for even $n$ and odd $n$ separately, then you get two geometric series.

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Your series is not an alternating one. You can rewrite it as $\sum_{n=1}^{\infty} (\frac{1}{5})^{2n-1} + (\frac{1}{3})^{2n} $ which should help you prove its convergence and find the limit

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  • $\begingroup$ Can you show the steps as to how you rewrote the series? Thanks! $\endgroup$
    – Jeel Shah
    Jun 28, 2016 at 14:44
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    $\begingroup$ Like the others said, separate odd and even terms. $\endgroup$
    – Vincent
    Jun 28, 2016 at 14:49
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HINT

Look at the first few terms of the series:

$$\frac{1}{(4-(-1))^n}, \frac{1}{(4-1))^n}, \frac{1}{(4-(-1))^n} \ldots$$

The above is for $n=1, 2, 3$. Do you notice a pattern? Can you see why it is not alternating?

As for convergence, the best (quite probably the simplest) option you have is comparison test.

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