1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The starting point 3 is irrelevant. $\endgroup$ – Did Feb 16 '14 at 15:55
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 !

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.