Does the sum of a converging and diverging series converge or diverge?

Question

Can a sum of two sequences – one that converges and one that diverges – result in a sequence that converges? My intuition says it cannot, but I don't see how it can be proved or disproved.

Answer 1

If sum of a convergent and a divergent sequence is convergent then we can write that divergent sequence as the sum(or subtraction) of two convergent sequence which is a contradiction.

Answer 2

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

Assume that $\sum_{n=1}^{\infty} (a_n + b_n)$ converges, but then $$\sum_{n=1}^{\infty} b_n = \sum_{n=1}^{\infty} (a_n + b_n) - \sum_{n=1}^{\infty} a_n$$and so $\sum_{n=1}^{\infty} b_n$ is the difference of two convergent series, so must converge.

So we have shown that $\sum_{n=1}^{\infty} b_n$ must converge if the sum converges.

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To show that the sum /difference of two convergent series is also convergent, just assume that $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ are convergent. Let $S_k = \sum_{n=1}^{k} a_n$ and $T_k = \sum_{n=1}^{k} b_n$. Then we know $S_k$ and $T_k$ converge as $k \rightarrow \infty$ (to say $S$ and $T$ respectively), so there is an integer $N>0$ large enough such that $$|S_k -S| < \frac{\varepsilon}{2}, \quad |T_k - T| < \frac{\varepsilon}{2}$$

Then by the triangle inequality, for $n>N$, we have $$\left|\sum_{n=1}^{k} (a_n + b_n) - (S + T) \right| \leq \left|\sum_{n=1}^{k} a_n - S \right| + \left|\sum_{n=1}^{k} b_n - T \right|$$ $$< \varepsilon$$ for any $\varepsilon >0$, which is exactly what it means to say that $\sum_{n=1}^{\infty} (a_n + b_n)$ converges to $S+T$