Let $\sum_{n=1}^\infty a_n$ be a positive convergent series and let $\sum_{n=1}^\infty b_n$ be a positive divergent series.

Prove that there exists an infinite number of n's such that $b_n\gt a_n$

Hey all. This is a very simple calculus problem, yet I am having troubles with formal proofs in this field.

As I see it, $\sum a_n$ is a positive convergent series, therefore the sequence $(a_n)$ is monotonic decreasing and $a_n\rightarrow 0$ as $n\rightarrow \infty$.

Let $S_n = \sum_{k=1}^{n} a_n$ and $T_n=\sum_{k=1}^{n} b_n$. There exists $L\in\mathbb{R}$ such that $\lim_{n\to\infty}S_n=L$ but $\lim_{n\to\infty}T_n=\infty$ $\Rightarrow T_n$ is unbounded.

Therefore for every $r\in\mathbb{R}$ there exists $N_0\in\mathbb{N}$ such that for all $n\ge N_0\rightarrow T_n\gt r$. In particular for $r=L$ there exists $N_0$ such that for all $n\ge N_0\rightarrow T_n \ge L$.

Hence $T_n=\sum_{k=1}^n b_k \gt \sum_{n=1}^\infty a_n = L$

Does that imply the required? I am not sure how to formally prove this and not sure if this direction is okay. I would love to hear your advices. Thank you!


Assume not. Then there are finitely many $n$'s so that $b_n>a_n$. Therefore there exists an $N$ so that for any $n>N$ we have $b_n\le a_n$. Now consider $S_n=\sum_{n=1}^{N}b_n$. Since $b_n\ge 0$ then $\{S_n\}$ is increasing. Also $$\lim_{n\to\infty}S_n=\sum_{n=1}^{\infty}b_n=\sum_{n=1}^{N}b_n+\sum_{n=N+1}^{\infty}b_n\le \sum_{n=1}^{N}b_n+\sum_{n=N+1}^{\infty}a_n$$ let $k=\sum_{n=1}^{N}b_n$ and $l=\sum_{n=N+1}^{\infty}a_n$ which are bounded. The former is a finite summation of bounded sequence and the latter one is convergent according to assumption. So:$$\lim_{n\to\infty}S_n\le l+k$$ Then the sequence $\{S_n\}$'s is an increasing upper bounded sequence then the limit exists and is bounded which implies on boundedness of $\sum_{n=1}^{\infty}b_n$ and is a contradiction. So we have proven what we want.


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