Prove the converges of∑n=1∞2n−1ln⁡(4n+1)n(n+1)

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Gene Espinosa · Accepted Answer

I would suggest you start with some rough hand-wavy approximations and then make it rigorous. For example, when n large,  2n−1n(n+1)≈2nn2=2n−32  and ln⁡(4n+1)∼ln⁡n  Lets

koffiejkl · Accepted Answer

First let us argue informally. Note that 2n−1 in the long run behaves like 2n. Since we won&#039;t worry about constants, let us say it behaves like n.The bottom behaves like n2, so 2n−1n(n+1) behaves like 1n32The term ln⁡(4n+1), in the long run, grows very slowly, more slowly than any power of n. So in the long run, in particular, it grows more slowly than n14. Thus, in the long run, our original expression goes to 0 faster than 1n54.But ∑1∞ converges. It follows that our original series converges.In essence, in 1n32, the exponent 32 is plenty big enough to ensure convergence. So we grab an n14 from it to &quot;kill&quot; the ln⁡(4n+1) term. That leaves us with 1n54, which still goes to 0 fast enough.Now it is time to turn the above into a more formal argument, say by using the Limit Comparison Test. Let an be the n-th term of our series. We show thatlimn→∞an1n54=0Since n54=n32n14, we want to show thatlimn→∞2n−1n32n(n+1)ln⁡(4n+1)n14=0We can use L&#039;Hospital&#039;s Rule to show that limn→∞ln⁡(4n+1)n14=0. Standard techniques show thatlimn→∞2n−1n32n(n+1)=2This completes the Limit Comparison argument.

Prove the converges of \sum_{n=1}^\infty\frac{\sqrt{2n-1}\ln(4n+1)}{n(n+1)}

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