Convergence or divergence of a series of the form $\sum \frac{a_n}{n^p}$

Key takeaway: For a series like $\sum_{n=1}^{\infty}\frac{a_n}{n^p}$, the answer depends on how the coefficients $a_n$ behave. The power $n^p$ helps, but it does not decide everything by itself.

Let

$\sum_{n=1}^{\infty}\frac{a_n}{n^p}$

be the series in question.

1) The first thing to check

You need information about $a_n$.

• If $a_n$ is bounded, then the series behaves like a $p$-series only after more detail is known.
• If $a_n \to 0$ slowly or does not approach $0$, convergence may fail.
• If $a_n$ has a limit $L \neq 0$, then the term is asymptotically like $\frac{L}{n^p}$.

2) Useful comparison

If $a_n \to L$ where $L\neq 0$, then

$\frac{a_n}{n^p} \sim \frac{L}{n^p}.$

So the series behaves like the $p$-series

$\sum \frac{1}{n^p},$

which:

• converges if $p>1$,
• diverges if $p\le 1$.

3) If $a_n$ is only bounded

If $|a_n|\le M$, then

$\left|\frac{a_n}{n^p}\right|\le \frac{M}{n^p}.$

So for $p>1$, the series converges absolutely by comparison. For $p\le 1$, boundedness alone is not enough to decide.

4) What you can say in an exam if no more data is given

Without conditions on $a_n$, the series is not determined. It may converge or diverge depending on the sequence.

For example:

• if $a_n=1$, then it becomes $\sum \frac1{n^p}$,
• if $a_n=n^p$, then every term is $1$ and the series diverges,
• if $a_n=(-1)^n$, the answer depends on $p$ and the convergence test used.

So the safest conclusion is: need more information about $a_n$.

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Pitfalls the pros know 👇 Don’t assume the denominator $n^p$ automatically forces convergence. The coefficient sequence matters. A second mistake is checking only termwise limit: even when $\frac{a_n}{n^p}\to 0$, the series may still diverge if the decay is too slow.

What if the problem changes? If the problem states that $a_n\to L$ with $L\neq 0$, then the series converges exactly when $p>1$. If $a_n$ is bounded and alternating, you may need the Alternating Series Test or Dirichlet-type ideas, depending on the exact form of $a_n$.

Tags: p-series, comparison test, absolute convergence