How to prove that a sequence converges

Question

I am having some trouble understanding how I can show that a given series converges.

I found a general explanation here that states:

To prove that a sequence converges, it is sometimes easier to start by finding a subsequence that converges (or proving that such a subsequence exists). Then one can hope to deduce that the sequence itself converges. Another way of using subsequences is to exploit the following result: if every subsequence has a further subsequence that converges to a limit L, then the whole sequence converges to L.

I was wondering whether somebody could explain this to me in more detail with an example? What's the trick to finding the sub-sequence?

Answer 1

Are you referring to series or sequences?

• If you're talking about sequences, you'll have in most cases to evaluate its limit for $n\to+\infty$. If for some reasons you can't, e.g. it's given by a recursion formula, then you could use the so-called Cauchy condition: if the sequence lies in a complete space (such as $\mathbb R$, $\mathbb C$), it converges if and only if it satisfies the condition that for all $\epsilon>0$, there exists an $N\in\mathbb N$ such that for all $n, m>N$ you have $|a_n-a_m|<\epsilon$.
• If you're talking about series, directly calculating its limit is not, generally, the easiest way to prove it converges. There are many other criteria, though, you can use to prove it without calculating the sum.

That explanation you gave with subsequences isn't usually the easiest (at least, as far as I know). If you can't guess the limit of the whole sequence, you can start with a simpler subsequence: this is just a hunch, though: a sequence converges to a limit $\ell$ if and only if each of its subsequences converges to $\ell$ too. You can see by yourself that this can't be used to determine the convergence of a sequence: you just can't calculate an infinite amount of limits! Anyway, if an easy subsequence you found converges to a certain limit, then you know what the overall limit should be. If that's not the case, then your sequence doesn't have a limit. (Note that this limit I'm speaking of might as well be infinite.)

Here's a not-so-trivial example: $a_n=\cos(\pi n)\frac1{n}$. Using subsequences, you can try the one in which $\cos(\pi n)$ is constant (it can be either $-1$, $0$ or $1$, as $n$ is an integer), so we'll take $\cos(\pi n)=1$, so $n=2k$ ($k\in\mathbb N$). Thus we have $a_{2k}=\cos(2k\pi)\frac1{2k}=\frac1{2k}$ which converges to zero. You should expect, then, that the overal limit of $a_n$ is zero too, but you can't use this result to prove that the limit is really zero (see next example), so you have still to use other methods.

An example of a sequence that doesn't have a limit: the sequence $b_n=(-1)^n$. You can find the subsequence for even values of $n$, that is (using $n=2k$, with $k\in\mathbb N$) $b_{2k}=(-1)^{2k}=1^k=1$, so it converges to $1$. But then you find another subsequence, for odd $n$ ($n=2k+1$), that is $b_{2k+1}=(-1)^{2k+1}=-1$, so those two have different limits: consequently $b_n$ cannot converge (in fact, it oscillates between $1$ and $-1$).

As for your question «What's the trick to finding the sub-sequence?»: intuition.

Answer 2

If you are talking about a sequence here is an example:

Show that if $x_{n}\rightarrow x$ and $|x_{n} - y_{n}| \rightarrow 0$ then $y_{n}\rightarrow x$.

proof: Let $\epsilon > 0$, since $x_{n}$ converges to $x$, then there exists a positive integer $n_{0}$ such that $$n > n_{0} \Rightarrow |x_{n} - x| < \epsilon$$ Also, since $|x_{n} - y_{n}|$ converges to $0$ then there also exists a positive integer $n_{1}$ such that $$n > n_{1} \Rightarrow |x_{n} - y_{n}| < \epsilon$$ So, now we can choose an $N\in \mathbb{N}$ such that $|x_{n} - x| < \epsilon/2 $ and $|x_{n} - y_{n}| < \epsilon/2 \ \ \forall n\geq N$. Hence by the Triangle Inequality, $\forall n\geq N$ $$|y_{n} - x| = |y_{n} - x_{n} + x_{n} - x| \leq |y_{n} - x_{n}| + |x_{n} - x| < \epsilon/2 + \epsilon/2 = \epsilon$$ Therefore, $y_{n} \rightarrow x$