# If $a_n$ is sequence of positive numbers such that $\sum a_n$ converges, then does $\sum \frac {{(a_n)}^{\frac 14}}{n^{\frac 45}}$ converge?

Failed attempts :

Let $x_n= \frac {{(a_n)}^{\frac 14}}{n^{\frac 45}}$.

1) By limit comparison test - Taking $y_n=a_n$, $\lim \frac {x_n}{y_n}=\lim \frac 1{n^{\frac 45} {a_n}^{\frac 34}}$ (Leads nowhere.)

Taking $y_n=\frac 1n$, $\lim \frac {x_n}{y_n}=\lim \frac 1{n^{\frac 15} {a_n}^{\frac 14}}$ (Leads nowhere).

Similarly taking $y_n={a_n}^2,\frac 1{n^2}$ doesn't work. Still searching for $y_n$ that works.

2) By comparison test - $\frac {{(a_n)}^{\frac 14}}{n^{\frac 45}} \le {(a_n)}^{\frac 14}$ (Leads nowhere since we can't say whether $\sum{a_n}^{\frac 14}$ converges).

3) By ratio test - $\lim\frac {{(a_{n+1})}^{\frac 14}}{{n+1}^{\frac 45}} \frac {n^{\frac 45}}{{a_n}^{\frac 14}}=\lim \frac {{a_{n+1}}^{\frac 14}}{{a_n}^{\frac 14}}$ (Leads nowhere since we don't know whether the last limit is less than $1$)

Where am I making mistakes? Can you provide some hints?

Hint:

For $u,v,w,t$ real and $\geq 0$, you have $$uvwt\leq \frac{u^4+v^4+w^4+t^4}{4}$$

Apply this with $u=a_n^{1/4}$, $\displaystyle v=w=t=\frac{1}{n^{4/15}}$

• How do you come up with using AM-GM? – Alex Vong May 12 '17 at 8:47

You can also use Holder's inequality: Take $p= 4, q=4/3,$ so that $1/p + 1/q=1.$ Holder then gives

$$\sum a_n^{1/4}n^{-4/5} \le \left [\sum (a_n^{1/4})^p)\right ]^{1/p}\cdot \left [\sum (n^{-4/5})^q \right ]^{1/q} = \left [\sum a_n \right ]^{1/4}\cdot \left [\sum n^{-16/15}\right ]^{3/4}.$$

On the right, the first factor is finite by assumption, and since $-16/15 < -1,$ the second factor is finite as well.

• I feel bad about myself that I didn't think about Holder's inequality. Such an easy proof. +1 – Error 404 May 12 '17 at 16:57
• @VikrantDesai Don't worry about it, we all miss things at times. I bet you'll think about Holder in the future in such situations. – zhw. May 12 '17 at 17:03
• Definitely! Thank you so much. :) – Error 404 May 12 '17 at 17:09