## September 1999 #1

Problem Statement: Let $\{a_n\}_{n=1}^{\infty}$ be a sequence of non-negative real numbers. Assume there exists a real number $k>1$ such that $\{n^{k}a_n\}_{n=1}^\infty$ converges. Show $\sum_{n=1}^\infty a_n$ converges.

Proof: Let $\{a_n\}$ be a sequence of non-negative real numbers. We know there exists a real number $k>1$ such that $\{n^k a_n\}$ converges. Since the sequence converges it is also Cauchy. Let $\lim_{n\rightarrow\infty}a_n=L$.

Let $\varepsilon>0$, then there exists $N'\in\mathbb{N}$ such that $|n^{k}a_{n}-L|<\varepsilon$ for every $n\geq N'$. This implies that for $n\geq N'$:

$L-\varepsilon

$\dfrac{L-\varepsilon}{n^k}

$0\leq |a_{n}|<\dfrac{L+\varepsilon}{n^k}$

As $n\rightarrow\infty$ it follows that $\dfrac{L+\varepsilon}{n^k}\rightarrow 0$. Since $0\leq |a_{n}|<\dfrac{L+\varepsilon}{n^k}$ for every $n\geq N'$ we may apply the Squeeze Theorem and conclude that $\{a_n\}$ converges to $0$.

Suppose that $\sum_{n=1}^\infty a_n$ diverges. Then there is a $N''\in\mathbb{N}$ such that $|S_{n+1}-S_n|>\varepsilon$ for every $n\geq N''$. Let $N=max\{N',N''\}$. Then for $n\geq N$ it follows that

$|S_{n+1}-S_n|>\varepsilon$

$|(a_1+a_2+\dots+a_{n+1})-(a_1+a_2+\dots+a_n)|>\varepsilon$

$|a_{n+1}|>\varepsilon$

But for $n\geq N$ it follows that $|a_{n+1}|<\varepsilon$ since $\{a_n\}\rightarrow 0$. Thus, $\varepsilon<|a_{n+1}|<\varepsilon$, which is a contradiction.Therefore we may conclude that $\sum_{n=1}^\infty a_n$ converges.

$\Box$

Reflection: When first attacking this proof we were getting stuck because we kept proving that $\{a_n\}$ converges to $0$, which ,as I’m sure we all know, tells us absolutely nothing about the series.

After completing this proof we consulted the “qual binder” and saw another, much much much simpler way to attack this problem. The argument follows from the fact that $\{a_n\}$ converges to $0$. Since the sequence converges we know that for $\varepsilon>0$ there is an $N\in\mathbb{N}$ such that $|n^{k}a_{n}-L|<\varepsilon$ for every $n\geq N$. This allows us to split the series into parts:

$\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{N}a_n+\sum_{n=N+1}^{\infty}a_n$

$<\sum_{n=1}^{N}+\sum_{n=N+1}^{\infty}\dfrac{L+\varepsilon}{n^k}$

Since $k>1$ we may conclude that $\sum_{n=N+1}^{\infty}\dfrac{L+\varepsilon}{n^k}$ converges by the p-series test and since it’s a finite sum we know that $\sum_{n=1}^{N}a_n$ converges. Thus, by the comparison test we may conclude that $\sum_{n=1}^{\infty}a_n$ converges. $\Box$