Question #13952

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Answer 1 · 2017-12-07T18:02:23

Diverges.

Rearranging a little:

$(\frac{1}{3}) \infty \sum k = 0 (5^{k})$ $(1/3)sum_(k=0)^oo(5^k)$

This is a geometric series with $r = 5$ $r=5$ . Since $r > 1$ $r>1$ we know that the series diverges.

Answer 2 · 2017-12-07T18:10:28

Divergent.

This is a geometric series with common ratio of $5$ $5$ .

The sum of a geometric series is given by:

$a (\frac{1 - r^{n}}{1 - r})$ $a((1-r^n)/(1-r))$

Where $a$ $a$ is the first term, $r$ $r$ is the common ratio and $n$ $n$ is the nth term.

If $| r | < 1$ $|r|<1$

Then the sum to infinity is:

$\frac{a}{1 - r}$ $a/(1-r)$

In this example:

The common ratio $r$ $r$ is $5$ $5$ :

$| 5 | > 1$ $|5|>1$

So the series diverges.