Example as this set of series do not converge

Question

Say you have a convergent series L = $\sum_{n=0}^\infty a_n < \infty $

Why is $\sum_{n=0}^\infty \frac{1}{a_n}$ not convergent? I mean say L $\in$ ℝ then shouldn't 1/L also be $\in$ ℝ? What steps to I have to make in my head to find this out by myself?

Thank you

Answer 1

You seem to be assuming that $\sum_{n=0}^{\infty}\frac{1}{a_n} = \frac{1}{\sum_{n=0}^{\infty}a_n}$, but this is completely false.

In general, $\frac{1}{a+b}\neq\frac1a+\frac1b$, and $\frac{1}{a+b+c}\neq\frac1a+\frac1b+\frac1c$, and so on. So this obviously won't work for infinite sums either.

In fact, if one of $\sum_{n=0}^{\infty}\frac{1}{a_n}$ and $\sum_{n=0}^{\infty}a_n$ converges to a finite value the other must diverge, because for a series to converge the terms have to converge to 0, and you can't have $a_n$ and $\frac{1}{a_n}$ both converging to 0.

Infinite summation is linear: If $\sum_{n=0}^{\infty}a_n=a$ then $\sum_{n=0}^{\infty}ca_n=ca$, and if also $\sum_{n=0}^{\infty}b_n=b$ then $\sum_{n=0}^{\infty}(a_n+b_n)=a+b$, but that's basically it. For an arbitrary function $f$ you don't have $\sum_{n=0}^{\infty}f(a_n)=f(a)$. For example, in general $\sum_{n=0}^{\infty}a_n^2\neq a^2$, for the same reason that in general $(a+b)^2\neq a^2+b^2$ and $(a+b+c)^2\neq a^2+b^2+c^2$.

If you're only interested in whether the series converges and not what is its sum, there are plenty of convergence tests to use, but here too you need to be careful and not make things up.

Answer 2

I am assuming that $(\forall n\in\mathbb{Z}^+):a_n\neq0$. Otherwise, the problem doesn't make sense.

If the series $\sum_{n=0}^\infty a_n$ converges, then $\lim_{n\to\infty}a_n=0$. Therefore, you don't have $\lim_{n\to\infty}\frac1{a_n}=0$ and so the series $\sum_{n=0}^\infty\frac1{a_n}$ diverges.

It seems, from you question, that you are mixing $\sum_{n=0}^\infty$ with $\lim_{n\to\infty}$.

Answer 3

I think the answer is simple. You have $\sum_{n=0}^\infty a_n $ converges. This implies $\lim_{n\to\infty}a_n=0$. And in this case $(1/a_n)_{n\to\infty}$ does not converge to zero. Hence $\sum_{n=0}^\infty \frac{1}{a_n}$ does not converge.

And to be more precise, you can take $a_n=\frac{1}{n^2}$ so $\sum_{n=1}^\infty \frac{1}{n^2} =\pi^2/6 $

and $\sum_{n=1}^\infty n^2$ is clearly divergent.