# Series Converging Almost Surely But Diverging in Mean

I am looking for an example of independent, non-negative random variables $$X_1, X_2, \dots$$ such that

$$\sum_{n=1}^{\infty} X_n \, \lt \, \infty$$

almost surely but

$$\sum_{n=1}^{\infty} \mathbb{E}(X_n) \, = \, \infty$$

I can find examples of sequences which converge almost surely but diverge in mean, but can’t seem to be able to cook up an example with a series.

## 1 Answer

For example, take $$X_n$$ s.t. $$P(X_n = 2^n) = \frac{1}{2^n}$$, $$P(X_n = 0) = 1 - \frac{1}{2^n}$$. Then a.s. all but finitely many $$X_n$$ are zeroes, so $$\sum X_n$$ converges a.s., but $$\mathbb E (X_n) = 1$$, so series of mean diverges.

• In your example of transforming a sequence example to a series example (i.e. $X_{n+1} = Y_{n+1} - X_n$), does this not result in random variables, $X_n$, which are not independent? – John Don May 25 '19 at 16:42
• Indeed it does. Probably I have missed the independence when thought about it. – mihaild May 25 '19 at 16:49