# Limit of $a_0 = 3 ; a_n = a_{n-1} + \frac{n-1}{n^2}$

I need to find the limit of $a_n$ for $n \rightarrow \infty$ but I am not sure how I would do it.

$$a_0 = 3 ; a_n = a_{n-1} + \frac{n-1}{n^2}$$

I tried to transform $a_n$ to a non recursive sequence but it never worked. For me it also looks more like a series than a sequence.

How would I tackle this problem?

My intuition tells me that it can not converge because $a_n$ starts with 3 and I always add 3 + something on the next term.

• The starting point 3 is irrelevant. – Did Feb 16 '14 at 15:55

## 5 Answers

Notice

\begin{align}&a_n = a_{n-1} + \frac{n-1}{n^2}\\ \implies &a_n -a_{n-1}=\frac{1}{n}-\frac{1}{n^2}\\ \implies \sum_{k=1}^{\infty}(&a_{k} - a_{k-1}) = \sum_{k=1}^{\infty}{\frac{1}{k}} - \sum_{k=1}^{\infty}{\frac{1}{k^2}}\\ \implies &a_{\infty} - a_0 = \sum_{k=1}^{\infty}{\frac{1}{k}} - \sum_{k=1}^{\infty}{\frac{1}{k^2}}\\ \implies &a_{\infty} = a_0 + \sum_{k=1}^{\infty}{\frac{1}{k}} - \sum_{k=1}^{\infty} {\frac{1}{k^2}}\end{align}

But we know that the harmonic series diverges while the sum of the reciprocals of all positive perfect squares converges (to $\frac{\pi^2}{6}$ in fact). Hence the series diverges.

Yes, it goes to infinity since your sequence is the sum of 3 and harmonic series which is infinity

It does not converge because $\displaystyle a_n = a_0 +\sum_{r=1}^{n}\left(\frac{1}{n}-\frac{1}{n^2}\right)$. Although $\sum\frac{1}{n^2}$ converge but $\sum\frac{1}{n}$ does not.

Your sequence is, as you rightly pointed out, a series. Your intuition, however, is wrong. For example, look at this sequence:

$$a_0=1$$ $$a_{n+1}=a_n + 2^{-n-1}$$

You can quickly see that $a_n = \sum_{i=0}^n \frac{1}{2^n}$, a sequence that converges, no even though, as you say, you always add something.

Look at your sequence this way:

$$a_n = a_{n-1} + \frac{n-1}{n^2} = \left(a_{n-2} + \frac{n-2}{(n-1)^2}\right) + \frac{n-1}{n^2}=\\ \left(\left(a_{n-3} + \frac{n-3}{(n-2)^2}\right)+ \frac{n-2}{(n-2)^2}\right) + \frac{n-1}{n^2}$$ If I continue this expansion, can you see that in fact, $a_n$ is actually a partial sum?

Your intuition is good and you have the proof : starting at $3$ and always add something which looks as the harmonic series (as already pointed out by Jlamprong).

Just for your curiosity, I give you here the closed form $$a(n)=3+\gamma-\frac{\pi ^2}{6} +\psi ^{(0)}(n+1)+\psi ^{(1)}(n+1)$$ where $\psi$ is the polygamma function and $\gamma$ the Euler constant. The limit of this is just $\infty$.

Added after 5xsum's answer

Suppose that we change $n^2$ by $n^3$ in your series; it converges to $3+\frac{\pi ^2}{6}-\zeta (3)$, while we still add something to $3$ as before !