Question

`\sum_(n=2)^\infty ((x+2)^n)/(2^n\lnn)`?

"Find the radius of convergence and interval of convergence."
11.8

`a_n=((x+2)^n)/(2^n\lnn)`
`a_(n+1)=((x+2)^(n+1))/(2^(n+1)\ln(n+1))=((x+2)^n(x+2))/(2^n(2)\ln(n+1))`

Applying Ratio Test:
`\lim_(n\rarr\infty)|a_(n+1)/a_n|=\stackrel(L)(\infty)|((x+2)\lnn)/(2\ln(n+1))|...`

What should I do next?
I know that after I find the ROC with `\abs(x)\lt1`, I should test the endpoints...
...but I can't even get those right now. I'm stuck on the Ratio Test.

Answer 1

`R=2,` interval of convergence: `[-4, 0)`

Explanation:

So, lifting off from where you ended, let's apply the Ratio Test:

`a_(n+1)=(x+2)^(n+1)/(2^(n+1)ln(n+1))`
`a_n=(x+2)^n/(2^nlnn)`

Thus,

`L=lim_(n->oo)|(x+2)^(n+1)/(2^(n+1)ln(n+1))*(2^nlnn)/(x+2)^n|`

`(x+2)^(n+1)/(x+2)^n=(x+2)`

`2^n/2^(n+1)=1/2`

Thus, factoring the `1/2` and `(x+2)` outside, maintaining its absolute value, we get

`1/2|x+2|lim_(n->oo)ln(n)/ln(n+1)` (We drop the absolute values on the logarithms as we're heading toward infinity -- everything is positive)

`lim_(n->oo)ln(n)/ln(n+1)=oo/oo` -- indeterminate, we should quickly apply l'Hospital's Rule, rewriting with a hypothetical differentiable variable `y`, as `n` is not differentiable:

`lim_(n->oo)lnn/ln(n+1)=lim_(y->oo)lny/ln(y+1)=lim_(y->oo)(1/y)/(1/(y+1))=lim_(y->oo)(y+1)/y=1`

Thus, we see `lim_(n->oo)lnn/ln(n+1)=1,` and we know we have convergence when `L<1` or

`1/2|x+2|<1`

`|x+2|<2 -> R=2`

Now, determine the interval:

`-2<x+2<2`

`-4<x<0`

Test these endpoints:

`x=0:`

`sum_(n=2)^oocancel(2^n)/(cancel(2^n)lnn)=sum_(n=2)^oo1/lnn`

We'll use the Direct Comparison Test: `1/n<=1/ln(n)` on `[2, oo)`, we know this because the logarithm grows slower, and the smaller denominator ensures a larger sequence, so, since the smaller series `sum_(n=2)^oo1/n` diverges by the `p-`series test where `p=1`, so does the larger series.

This endpoint is thus excluded from the interval of convergence.

`x=-4:`

`sum_(n=2)^oo(-2)^n/(2^nlnn)=sum_(n=2)^oo((-1)^ncancel(2^n))/(cancel(2^n)lnn)=sum_(n=0)^oo(-1)^n/lnn`

Using the Alternating Series Test, we see `b_n=1/lnn` is decreasing on `[2, oo)` due to the growing denominator and `lim_(n->oo)1/lnn=0`

So, for this endpoint, we have convergence by the Alternating Series Test.

Interval of Convergence: `[-4, 0)`

Answer 2

The interval of convergence is `-4<=x<0`

Explanation:

`lim_(n->+oo)|a_(n+1)/a_n|=lim_(n->+oo)|((x+2)^(n+1)/(2^(n+1)ln(n+1)))/((x+2)^(n)/(2^(n)ln(n)))|`

`=lim_(n->+oo)|(x+2)/2lnn/(ln (n+1))|`

`=|(x+2)/2|` as `lim_(n->+oo)|lnn/(ln (n+1))|=1`

The series converges for

`|(x+2)/2|<1`

`-2 < x+2 <2`

`-4 < x <0`

For `x=0`

`sum_(n=2)^oo(x+2)^(n)/(2^(n)ln(n))=sum_(n=2)^oo2^n/(2^2lnn)=sum_(n=2)^oo1/lnn`

The series diverges for `x=0`

For `x=-4`

`sum_(n=2)^oo(-4+2)^(n)/(2^(n)ln(n))=sum_(n=2)^oo(-2)^n/(2^2lnn)=sum_(n=2)^oo (-1)^n/(lnn)`

This series converges conditionally for `x=-4`

The interval of convergence is `-4<=x<0`