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Please, give me an example where $X(n)$ and $Y(n)$ are both divergent series, but $(X(n) + Y(n))$ converges.

Walter r
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4 Answers4

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How about $X_n = -n$ and $Y_n = n$ for $n \geq 1$.

Lord Soth
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Try $x_n = n, y_n = -n$. Then both $x_n,y_n$ clearly diverge, but $x_n+y_n = 0$ clearly converges.

Or try $x_n = n, y_n = \frac{1}{n}-n$ if you want something less trivial. Again both $x_n,y_n$ diverge, but $x_n+y_n = \frac{1}{n}$ converges.

copper.hat
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Take $y_n = -x_n$ for some divergent $x_n$ since $y_n$ will be divergent too.

Halil Duru
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As others have already pointed out, if $X(n)$ and $Y(n)$ diverge, this need not always mean that $X(n) + Y(n)$ diverges.

A slightly general example is of the following form.

Consider a convergent sequence $c(n) \to c \in \mathbb{R}$. Now pick any divergent sequence $X(n)$. Consider the sequence $Y(n) = -X(n) + c(n)$.

Then we have that if $X(n) \to \infty$, then $Y(n) \to -\infty$ and if $X(n) \to -\infty$, then $Y(n) \to \infty$. Finally, $X(n) + Y(n) \to c \in \mathbb{R}$ and hence it converges.

However, note that if we want $X(n)$ and $Y(n)$ to diverge, but $X(n) + Y(n)$ to converge, then one of them must diverge to $+ \infty$ and the other must diverge to $-\infty$.

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    Your last argument does not appear to be correct. How about $X_n = (-1)^n$ and $Y_n = -X_n$? – Lord Soth Apr 08 '13 at 18:15
  • @LordSoth By divergent, I assume it doesn't oscillate, i.e., it either tends to $+\infty$ or $-\infty$. –  Apr 08 '13 at 18:17