Can Divergent Series Sums Converge?

[PFuser1232]

Consider the two divergent series:
$$\sum_{n=k}^{\infty} a_n$$
$$\sum_{n=k}^{\infty} b_n$$
Is it possible for $\sum_{n=k}^{\infty} (a_n \pm b_n)$ to converge?

 

[Orodruin]

Yes, consider the case $a_n = b_n$. Then the sum of $a_n - b_n$ is zero.

 

[PFuser1232]

Yes, consider the case $a_n = b_n$. Then the sum of $a_n - b_n$ is zero.

Thanks!
But in that case does it make sense to say that $\sum_{n=k}^{\infty} (a_n \pm b_n) = \sum_{n=k}^{\infty} a_n \pm \sum_{n=k}^{\infty} b_n$?

 

[Orodruin]

Thanks!
But in that case does it make sense to say that $\sum_{n=k}^{\infty} (a_n \pm b_n) = \sum_{n=k}^{\infty} a_n \pm \sum_{n=k}^{\infty} b_n$?

No, it generally make sense only if the series converge.

 

[PeroK]

Consider the two divergent series:
$$\sum_{n=k}^{\infty} a_n$$
$$\sum_{n=k}^{\infty} b_n$$
Is it possible for $\sum_{n=k}^{\infty} (a_n \pm b_n)$ to converge?

You really ought to be able to find examples yourself to resolve this question.

 

[Ssnow]

You can construct others, for example $a_ {n}=\frac{1}{n^{2}}+b_{n}$ so $\sum_{n=1}^{\infty}a_{n}-b_{n}=\frac{\pi^{2}}{6}$...