Let $\{a_n\}$ be a sequence of non-negative real numbers satisfying $a_{n+1} \leq a_n + \frac{(-1)^n}{n}$. Prove $\{a_n\}$ converges.

It is easy to show $a_n$ is bounded by $a_1$ plus the alternating harmonic series. By similar means, it is easy to provide a small upperbound on the difference $a_m - a_n$ for $m > n$. However, I'm having difficulty giving a lower bound for this difference to give us an absolute bound. I'm now searching for alternative approaches.

  • $\begingroup$ You can use non-negativity of the $a_i$ values to get a lower bound. Even/odd observations may also help. $\endgroup$ – Michael Sep 19 '17 at 1:33

Probably it is easier to show $\{a_n\}$ converges by showing that $\limsup_{n\to\infty}a_n = \liminf_{n\to\infty}a_n$. I'll give a sketch below: I think you'll be able to fill in the details.

Since $a_n$ consists of nonnegative real terms, the $\liminf$ is finite, so for every $\epsilon > 0$ and for every $N$, you can find an $n>N$ such that $a_n - \liminf_{n\to\infty} a_n < \frac{\epsilon}{2}$. Also, by the convergence theory for alternating series, you can choose an $N$ such that $\left|\sum_{k=n}^m \frac{(-1)^k}{k}\right| < \frac{\epsilon}{2}$ for all $n > N$.

Now, using your reasoning, $a_m \le a_n + \left|\sum_{k=n}^m \frac{(-1)^k}{k}\right| < \liminf_{n\to\infty} a_n + \epsilon$. Taking the $\limsup$ now and since $\epsilon > 0$ was arbitrary, you can find that $\limsup_{n\to\infty}a_n \le \liminf_{n\to\infty} a_n$. This implies that the $\limsup$ and $\liminf$ are equal, so $a_n$ converges.

  • $\begingroup$ @Strants from $a_n- \liminf a_n<\epsilon$ it is proved that just exists a subsequence that converges to liminf and then that exists a convergent subsequence...i do not understand your solution.. $\endgroup$ – Marios Gretsas Sep 19 '17 at 2:32
  • $\begingroup$ @MariosGretsas Certainly, the fact that $a_n - \liminf a_n < \epsilon/2$ alone only proves the existence of a convergent subsequence. However, we can combine this fact with the bound $a_{n+1} \leq a_n + \frac{(-1)^n}{n}$ to show that no other subsequence can converge to anything greater than $\liminf a_n + \epsilon$. Since $\epsilon > 0$ was arbitrary, this shows that every subsequence (so in particular the sequence itself) converges to $\liminf a_n$. $\endgroup$ – Strants Sep 19 '17 at 2:42

Define $s_n = \sum_{k=1}^n \frac{(-1)^{n}}{n}$, and then define $b_n = a_{n+1}-s_n$. Note that $s_n$ is bounded (because it converges), and $a_n$ is also bounded (since $0 \leq a_n \leq a_1+s_n$). Therefore $b_n$ is bounded.

Note that $b_n = a_{n+1}-s_n \leq a_n +\frac{(-1)^n}{n}-s_n = a_n-s_{n-1}=b_{n-1}$. Thus $b_n$ is decreasing.

Since $b_n$ is bounded and decreasing, it converges. Since $b_n$ and $s_n$ are convergent, their sum $a_{n+1}=b_n+s_n$ also converges. Thus $a_n$ converges.

  • 1
    $\begingroup$ Nice. I like this one too. $\endgroup$ – GillyB Sep 19 '17 at 2:25

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