Example of Series such that $\sum a_n$ converges but $\sum a_n^4$ diverges

examples-counterexamplesreal-analysissequences-and-series

I am interested in finding

I am able to find converse like if $\sum a_n^4$ converges but $\sum a_n$ diverges using harmonic series

But I not able to find suitable example for above

Any help will be appreciated

Best Answer

Hint: Try something alternating, e.g. $$ a_n = \frac{(-1)^n}{\sqrt[4]{n}}. $$

Related Solutions

[Math] A series such that $\sum {a_n}$ converges, but $\sum {a_{3n}}$ diverges.

It is entirely fine to define a sequence term by term, and your examples are fine. In fact $\LaTeX$ even supports this with the following environment:

a_n=
\begin{cases}
 [value 1] & [condition 1] \\
 [value 2] & [condition 2] \\
 ...
\end{cases}

For example (right click to show underlying code): $$a_n= \begin{cases} 2&\text{ if }\ 3\text{ divides } n\\ -1&\text{ otherwise} \end{cases} \qquad\text{ and }\qquad b_n= \begin{cases} 0&\text{ if }\ 3\text{ divides } n\\ 1&\text{ otherwise} \end{cases}.$$

Let $a_n$ be an unbounded non decreasing sequence s.t. $\sum \frac{a_n}{n^a}$ diverges, for $a > 2$. Does $\sum 1/a_n$ converge

Its false, here is a counterexample. We define $a_n$ inductively. Start with $a_0=1$ and start with auxiliary sequence $k(n)$ with $k(0)=0$. Now:

If $k(n-1)=0$ set $a_n=2^n$ and $k(n)=2^n$. Else set $a_n=a_{n-1}$ and $k(n)=k(n-1)-1$.

Note that $k(n)=0$ will happen for infinitely many $n$, and for any $n$ where this happens you have $a_n=2^n$, so $\frac{a_n}{n^a}$ will be unboundedly large for such $n$, in particular the series over this has no hope of converging.

On the other hand if $n$ is so that $k(n)=0$ you have for all $m\in[n, n+2^n]$ that $a_m=2^n$, so: $$\sum_{m=n}^{n+2^n}\frac1{a_m}=2^n\cdot\frac1{2^n}=1$$ and you see that also the series $\sum \frac{1}{a_n}$ cannot converge.

Best Answer

Related Solutions

[Math] A series such that $\sum {a_n}$ converges, but $\sum {a_{3n}}$ diverges.

Let $a_n$ be an unbounded non decreasing sequence s.t. $\sum \frac{a_n}{n^a}$ diverges, for $a > 2$. Does $\sum 1/a_n$ converge

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