Suppose: <munderover> &#x2211;<!-- ∑ --> <mrow class="MJX-TeXAtom-ORD"> n =

Marianna Stone

Marianna Stone

Answered question

2022-05-22

Suppose:
n = 2 ( 1 n ( ln ( n ) ) k ) = 1 2 ( ln ( 2 ) ) k + 1 3 ( ln ( 3 ) ) k + . . . ,

Answer & Explanation

Erick Blake

Erick Blake

Beginner2022-05-23Added 10 answers

For completeness, we sketch the Integral Test approach.
Let f be a function which is defined, non-negative, and decreasing (or at least non-increasing from some point a on. Then 1 f ( n ) converges if and only if the integral a f ( x ) d x converges.
In our example, we use f ( x ) = 1 x ln k ( x ) . Note that f ( x ) 0 after a while, and decreasing So we want to find the values of k for which
2 d x x ln k ( x ) ( )
converges (we could replace 2 by say 47 if f(x) misbehaved early on).
Let I ( M ) = 0 M f ( x ) d x. We want to find the values of k for which lim M I ( M ) exists.
Suppose first that k > 1. To evaluate the integral, make the substitution ln x = u. Then
I ( M ) = 2 M d x x ln k ( x ) = ln 2 ln M d u u k .
We find that
I ( M ) = 1 k 1 ( 1 ( ln 2 ) k 1 1 ( ln M ) k 1 ) .
Because k 1 > 0, the term in M approaches 0 as M , so the integral (∗) converges.
By the Integral Test, we therefore conclude that our original series converges if k > 1
For k = 1, after the u-substitution, we end up wanting d u u . We find that
I ( M ) = ln ( ln M ) ln ( ln 2 ) .
As M (though glacially slowly). So by the Integral Test, our series diverges if k = 1
For k < 1, we could again do the integration. But an easy Comparison with the case k = 1 proves divergence.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?