Problem while proving the convergence $\sum \frac 1{n(n+1)^2}$ using Comparison Test.

Question

I found a test going by the name, "Comparison Test" to test convergence of a series.

It states that :

Let $ \sum u_n$ and $\sum v_n$ be two series of positive real numbers and there is a natural number $m$ such that $u_n\leq kv_n,\forall n\geq m,$ $k$ being a fixed positive number. Then, $\sum u_n$ is convergent if $\sum v_n$ is convergent.

Now, I know $\sum \frac 1{n(n+1)^2}$ is convergent. But there's a problem.

When I use the Comparison Test it turns out that $\frac{1}{(n(n+1)^2}\leq \frac 1{n^3}\leq \frac 1n$ for all natural numbers. Now, if $v_n=\frac 1n$ then, we know $\sum \frac 1n(=\sum v_n)$ is a harmonic series so, divergent which implies, $\sum \frac 1{n(n+1)^2}$ is divergent by Comparison Test.

Now, what's wrong with this argument?

Answer 1

Direct comparison test is used in this way, given that eventually

$$0\le a_n \le b_n \le c_n$$

then

$$\sum b_n \; \text{diverges} \implies \sum c_n \; \text{diverges}$$

but we can't say anything about $\sum a_n$ and

$$\sum b_n \; \text{converges} \implies \sum a_n \; \text{converges}$$

but we can't say anything about $\sum c_n$.

In your case the useful inequality to prove convergence is

$$\frac{1}{n(n+1)^2}\leq \frac 1{n^3}$$

and we don't need more than that.

Answer 2

No, it is not, because if you take $\sum\frac1n$ to compare with, you only know

$$\sum\frac1{n(n+1)^2}<\sum\frac1n=\infty$$

you can't say anything about it is convergent or divergent. For example, we know

$$\sum 0 <\sum\frac1n$$

and $\sum 0=0$ is convergent.

$$\sum \frac1{2n} <\sum\frac1n$$

but $\sum \frac1{2n}$ is divergent.

Answer 3

Your statement for comparison test is wrong. For two Positive sequences $x_n$ and $y_n$ such as $$x_n\leq y_n, $$ you can conclude that $\sum x_n$ converges if $\sum y_n$ does, and we can conclude that $\sum y_n$ diverges if $\sum x_n$ does. What really makes sense if you think about it.