Knowing that sum from n=1 to infinity 1/(n^()alpha)
Knowing that $\sum_{n=1}^{\infty}\frac{1}{n^\alpha}$ converges for $\alpha>1$ and diverges for $\alpha\le1$, determine the convergence of: $$ \sum_{n=1}^{\infty} \frac{1}{(an+b)^p}, \qquad a>0. $$
Explanation of Method
Since $a>0$, the linear expression $an+b$ grows proportionally to $n$ — it is just a rescaled and shifted version of $n$ itself. Such a linear change of variable never alters the convergence behavior of a series (only shifts finitely many terms and rescales the sum), so $\sum 1/(an+b)^p$ converges exactly when the ordinary $p$-series $\sum1/n^p$ does: for $p>1$.
- Identify the asymptotic behavior $an+b\sim an$.
- Apply the limit comparison test against $\sum1/n^p$.
- State the resulting convergence criterion in terms of $p$.
Step 1 · Asymptotic behavior
Since $a>0$, for large $n$: $an+b = n\left(a+\frac bn\right)\approx an$, so $a_n=\dfrac1{(an+b)^p}\approx\dfrac1{a^pn^p}$.
Step 2 · Limit comparison test
With $b_n=\dfrac1{n^p}$:
$$ \lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty}\left(\frac{n}{an+b}\right)^p = \left(\lim_{n\to\infty}\frac1{a+\frac bn}\right)^p = \frac{1}{a^p}. $$
Since $a>0$, this limit is positive and finite, so $\sum\frac1{(an+b)^p}$ and $\sum\frac1{n^p}$ share the same convergence behavior.
Step 3 · State the criterion
Since $\sum\frac1{n^p}$ converges iff $p>1$:
$$ \sum_{n=1}^{\infty}\frac1{(an+b)^p} \quad\text{converges iff } p>1,\ \text{diverges iff } p\le1. $$
This holds regardless of the specific values of $a>0$ and $b$ (any finitely many non-positive terms, if $an+b\le0$ for small $n$, may simply be dropped without affecting convergence).
Summary Table
| Quantity | Value |
|---|---|
| Asymptotic behavior | $\dfrac1{(an+b)^p}\sim\dfrac1{a^pn^p}$ |
| Limit comparison ratio (vs. $1/n^p$) | $\dfrac1{a^p}$ |
| Convergence | iff $p>1$ |
Additional Notes
- This result formalizes the intuitive fact that linear reindexing ($n\mapsto an+b$) never changes convergence: it only rescales the "speed" at which the index runs, and comparison tests are insensitive to constant rescalings.
- The parameter $b$ plays no role whatsoever in the convergence criterion — only $a>0$ (needed so the terms grow, not shrink or oscillate) and the exponent $p$ matter.
- Special case $a=1,b=0$ recovers the ordinary $p$-series exactly, confirming the general formula is consistent with the base case it generalizes.
Result
The convergence of the series depends only on $p$: