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Give an example of two divergent series of real numbers sch that their sum is convergent.

I have read that the sum of two divergent series can be divergent or convergent.

I have found that, the series $\sum_{n=1}^\infty\frac{1}{n}$ and $\sum_{n=1}^\infty\frac{1}{n+1}$ both are divergent series and their sum $\sum(\frac{1}{n}+\frac{1}{n+1})$ is also a divergent series.

Again, If we take $u_n=(-1)^n$ and $v_n=(-1)^{n+1}$
Then both $\sum u_n$ and $\sum v_n$ are divergent (Oscilatory). But their sum, i.e, $\sum (u_n+v_n)$ is convergent and equals to $0$.

But I cannot find any other example of two divergent series with their sum convergent.
How can I give such an example.
Please help...!!

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  • $\begingroup$ $$\sum\left[ (-1)^n+ (-1)^{n+1}\right] = \sum 0 = 0.$$ $\endgroup$ – Jack D'Aurizio Jul 27 '16 at 15:40
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How about $$ 1 + 2 + 3 + \ldots $$

and

$$ -1 + (-2) + (-3) + \ldots $$

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  • $\begingroup$ oh yeah....Thanks.. $\endgroup$ – Indrajit Ghosh Jul 27 '16 at 14:46
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$$\sum_n \frac{1}{n} +\sum_n \frac{-1}{n}$$

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This might be a bit trivial but : $$ A = \sum_{i=1}^{\infty} \frac{1}{i} \\ B = \sum_{i=1}^{\infty} -\frac{1}{i} \\ $$ $A+B = 0$ while $A = +\infty$ and $B = -\infty$

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Your question lacks a key piece of information. You are dealing with what is called a conditionally convergent series. https://en.wikipedia.org/wiki/Conditional_convergence

Basically, we say that a series is conditionally convergent if it converges, but the series formed by taking the absolute value of each value in the series does not converge. Otherwise, we say that it converges absolutely.

So, for example:

1 + 1/2 + 1/4 + 1/8 + ... converges absolutely to 2.

1 + 1/2 + 1/3 + 1/4 + 1/5 + ... diverges.

But, 1 - 1/2 + 1/3 - 1/4 + 1/5 + ... converges conditionally.

It has been proven that a conditionally convergent series may be rearranged to converge to any number, or my be rearranged to be divergent.

Therefore, your question does not make sense without defining how you take the sum of two series. If you were more specific and said that you would interweave the elements of the series one at a time, some of the examples give in other answers make sense. If you were to say that you simply add the resulting sums of the two series, then none of the examples make sense.

In short, it depends upon what order you add the values of the resulting series.

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There are also:

$$A=\sum_{x=1}^{\infty} \sin(x) $$ $$B=\sum_{x=1}^{\infty} \sin(x-\pi) $$

In general two sinusoidal waves that are out of phase have this propriety. Notice that in your case:

$$(-1)^n=\cos(\pi n)+i\sin(\pi n)$$ $$(-1)^{n+1}=\cos(\pi (n+1))+i\sin(\pi (n+1))$$

That are out of phase :) .

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$\sum_{0}^{\infty} \frac{1}{n+1}$ Diverges

$\sum_{0}^{\infty}\frac{-1}{n+2}$ Diverges

And$$\sum_{0}^{\infty}\bigg(\frac{1}{n+1}-\frac{1}{n+2}\bigg)$$ this converges, as the k-th partial term tends to a finite value as k tends to infinity

As $S_k=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{k+1}-\frac{1}{k+2}=1-\frac{1}{k+2} \rightarrow1$ Where $$S_{k}=\sum_{0}^{k}\bigg(\frac{1}{n+1}-\frac{1}{n+2}\bigg)$$

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