# Converging series

1. Jan 9, 2009

1. The problem statement, all variables and given/known data

Show that the following series converges:

2. Relevant equations

Sum of n=1 to n=infinity of [(n+1)/(n^2 +1)]^2

3. The attempt at a solution

I thought about using the ratio test, so if the limit as n tends to infinity of a(n+1) / a(n) < 1, then a(n) - the series - would converge.
Once i put a(n+1)/a(n), i don't really know how to continue.

2. Jan 9, 2009

### sutupidmath

$$\sum_{n=1}^{\infty}\frac{n+1}{(n^2+1)^2}$$

I would try to use the comparison test to establish the convergence of this series. My first guess is that it is convergent, so i will try to compare it to another convergent series.

I would start off by noticing that:

$$n^3\leq n^4$$

THen it is clear that:

$$n^3<n^4+2n^2+1=(n^2+1)^2$$

One more step towards the answer:

Taking the reciprocal of the above we get:

$$\frac{1}{n^3}>\frac{1}{(n^2+1)^2}$$

Now, all u need to do is multiply both sides of the inequality by n+1, try to manipulate a little bit the left side and justify ur answer.

I will stop here, and see whether u can get it done?

Last edited: Jan 9, 2009
3. Jan 9, 2009

i know for the comparison test, you assume that a(n)<= k.b(n)
Then if b(n) is convergent, a(n) is also convergent.

However i still dont see how i can get my series to be <= to a constant times another series.

4. Jan 9, 2009

### sutupidmath

The comparison test says: Let $$\sum b_n,\sum a_n$$ be two series with positive terms.
(i) If $$\sum b_n$$ is convergent, and if $$a_n<b_n$$ for some n>N, then the series $$\sum a_n$$ also converges.
(ii)If $$\sum b_n$$ divergent, and if $$b_n<a_n$$ for some n>N, then $$\sum a_n$$ is also divergent.

But, what u are saying there is also true, since if $$\sum b_n$$ converges, so does the series $$\sum k*b_n$$ where k is a constant.

5. Jan 9, 2009

### sutupidmath

As for this one, i just showed you in my 1st post.

6. Jan 10, 2009

### Gib Z

We can't use sutupidmath's particular example for this one, as its the entire expression that is squared, not just the denominator =[

However the problem is still easily tackled. First, lets consider the expression under the squared sign. That expression is larger than the analoguous expression with only n in the numerator, ie the +1 omitted. Then, dividing all terms by n, we need that expression is $1/( n + 1/n)$

After we square that, it becomes $1/ (n^2 + f(n))$ where f(n) are some positive terms. Hence, this expression is in turn larger than the series for $1/n^2$. And we know that converges.

7. Jan 10, 2009

### sutupidmath

Dang, i didn't see that!

8. Jan 10, 2009

### HallsofIvy

Staff Emeritus
with
$$\left[\frac{n+1}{n^2+1}\right]^2$$
You can still do the same thing. Divide both numerator and denominator inside the square by n2 and you get
$$\left[\frac{\frac{1}{n}+ \frac{1}{n^2}}{1+ \frac{1}{n^2}}\right]^2$$
Now it should be clear that, while the numerator goes to 0 as n goes to infinity, the denominator goes to 1. The fraction goes to 0 and, because x2 is continuous, so does its square.

9. Jan 11, 2009

### Дьявол

$$\lim_{n \rightarrow \infty}(\frac{n+1}{(n^2+1)^2})=0$$, so the series are convergent. If it wasn't equal to zero, than the series are divergent.

Regards.

10. Jan 11, 2009

### sutupidmath

Well, this is not quite true.

A necessary but not sufficient condition for a series $$\sum a_n$$ to converge, is if $$\lim_{n\rightarrow \infty}a_n=0$$

A simple example is the harmonic series $$\sum\frac{1}{n}$$

1/n -->0 as n-->infinity, but we know that the harmonic series diverges.

11. Jan 18, 2009

### Appa

Doesn't that only show that the sequence of terms converges? And that's not a sufficient condition for the series to converge too... To show that the series converges, shouldn't you show that the sequence of partial sums converges, i.e. it is bounded? I'm not sure how to do that in this case though.
I'm having trouble with the convergence of series myself, so I'm only asking.

12. Jan 18, 2009

### HallsofIvy

Staff Emeritus
Yes, but my point was that whatever "comparison" you get for $(n+1)/(n^2+1)$, compare $[(n+2)/(n^2+1)]^2$ to its square.

Certainly $(n+1)/(n^2+1)< 2n/(n^2+1)$ and it is not too hard to show that $n/(n^2+1)$ itself is less than 2/n. That itself would not be enough to show that $(n+1)/(n^2+1)$ itself converges because 2/n does not converge. But it does show that $[(n+1)/(n^2+1)]^2< 4/n^2$ and $4/n^2$ does converge.