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I was faced with two problems:

$$\sum_{n=1}^{\infty} \frac{ \cos^{2}(n) }{ n^{\frac{3}{2}} }$$

and

$$\sum_{n=1}^{\infty} \frac{ \tan^{-1}(n) }{ n\sqrt{n} }$$

I was lost when trying to find a way to compare these using the direct-comparison test and the limit comparison test. I wasn't sure how this was suppose to be set-up.

Is the idea to use the maximum of what each function can get on its domain, so for sine it's maximum height is $1$, and for $\tan^{-1}(n)$ it's maximum height is $\frac{\pi}{2}$.

I don't understand why we do this though. Why use it's maximums in the numerator of the comparable functions?

I used an online calculator that set up the comparison tests for these two series like so:

$$\sum_{n=1}^{\infty} \frac{ \cos^{2}(n) }{ n^{\frac{3}{2}} } < \sum_{n=1}^{\infty} \frac{ 1 }{ n^{\frac{3}{2}} }$$

$$\sum_{n=1}^{\infty} \frac{ \tan^{-1}(n) }{ n\sqrt{n} } < \sum_{n=1}^{\infty} \frac{ \frac{\pi}{2} }{ n\sqrt{n} }$$

I have no idea why we are using their maximum heights, or why this would make them greater than the originals. Is it because the sum for the originals would dip between it's entire domain instead of exclusively being it's maximum? Is that the intuition? But wouldn't that mean it's a bigger number being divided up making it smaller?

Very confused here

Please help

Thank you

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  • $\begingroup$ They are both bounded below by 0. If something that bounds the function from above converges it must converge. It's like a weaker version than the squeeze theorem. $\endgroup$ – mathreadler Apr 26 '17 at 18:12
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The Comparison Test: Suppose $\sum_nB_n$ converges and $B_n\geq 0$ for all $n.$ If there exists $K>0$ such that $|A_n|\leq KB_n$ for all but finitely many $n,$ then $\sum_nA_n$ also converges. Proof: Let $n_0$ be such that $n>n_0\implies |A_n|\leq KB_n.$ Then $$0\leq \lim_{n\to \infty}\sup_{n\leq m\leq m'}|\sum_{j=m}^{j=m'}A_j|=\lim_{n_0<n\to \infty}\sup_{n\leq m\leq m'}|\sum_{j=m}^{j=m'}A_j|\leq$$ $$ \leq \lim_{n_0<n\to \infty}\sup_{n\leq m\leq m'}\sum_{j=m}^{j=m'}|A_j|\leq$$ $$ \leq \lim_{n_0<n\to \infty} \sup_{n\leq m\leq m'}\sum_{j=m}^{m'}KB_n=0.$$ Since $K>0$, and since $B_n\geq 0$ for all $n$, the last equality, above, ("$=0$"), is equivalent to the convergence of $\sum_nB_n$.

In your Q , with $B_n=1/n\sqrt n,$ which converges by the Cauchy Condensation Test (and also by the Monotone Integral Test): For the first series let $K=1,$ and for the second series let $K=\pi /2.$ In each case we have $|A_n|\leq B_n$ for all $n.$

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