00:01
In this problem, we are provided with the series sigma n equals to 2 up to infinity, x raised to the power 3n plus 3, the whole divided by natural log of n.
00:17
We are asked to find out the interval of convergence.
00:26
So for finding out the interval of convergence, we make use of the ratio test.
00:32
So according to the ratio test, we must find out limit n tends to infinity, a .n plus 1 over a .n.
00:42
And if this value is less than 1, it implies that the series converges.
00:49
If the value is greater than 1, then it implies that the series diverges.
00:55
And if the value equals to 1, then it implies that the test is inconclusive.
01:04
So here let us first consider a .n.
01:07
A .n equals to x raised to the power 3n plus 3 over natural log of n.
01:14
Next we consider a .n plus 1.
01:16
A .n plus 1 equals to x raised to the power 3n plus 6.
01:21
The whole divided by natural log of n plus 1.
01:25
So now we can take the ratio...