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# Determine whether the series converges or diverges.$\displaystyle \sum_{n = 1}^{\infty} \frac {6^n}{5^n - 1}$

## Diverges

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were given a series and were asked to determine whether this series converges or diverges series is the sum from n equals one to infinity of six to the end, over five to the end, minus one. Well, we know that this is a some with strictly positive terms and so we know that six to the end over five to the n minus one. This is also strictly greater than six to the end over five to the end for all natural numbers and therefore it follows the some itself is greater than the sum from n equals one to infinity of six to the end over five to the end. Now, this can also be written as 6/5 times the sum from and equals one to Infinity of six fits to the n minus first power. However, you want to look at it. Either way, we see that this is a geometric series with common ratio are equal to six bits. Of course, the absolute value of R is greater than one, so the geometric series diverges series on the right. Now, by the comparison test, a series greater than a divergent series also diverges. Therefore, the given series diverges as well, and for the comparison, we used a geometric series

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