#\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.
"Find the radius of convergence and interval of convergence."
11.8
Applying Ratio Test:
What should I do next?
I know that after I find the ROC with
...but I can't even get those right now. I'm stuck on the Ratio Test.
2 Answers
Explanation:
So, lifting off from where you ended, let's apply the Ratio Test:
Thus,
Thus, factoring the
Thus, we see
Now, determine the interval:
Test these endpoints:
We'll use the Direct Comparison Test:
This endpoint is thus excluded from the interval of convergence.
Using the Alternating Series Test, we see
So, for this endpoint, we have convergence by the Alternating Series Test.
Interval of Convergence:
The interval of convergence is
Explanation:
The series converges for
For
The series diverges for
For
This series converges conditionally for
The interval of convergence is