Proving Convergence of the Sum for Real Parameter a in (-1,1)"

[twoflower]

Hi all,

my task is to solve the convergence of the sum in dependence to the parameter a real.

$$\sum_{n=1}^{\infty} \frac{a^{n}}{1+a^{n}}$$

I did it this way:

First I found out that if the sum converges, a will have to be in interval (-1, 1). So how to prove that for these values of parameter the sum converges?

Let's try d'Alembert's criterion, which tells us this:

$$\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}} < 1 \Rightarrow \sum a_{n}{} is convergent$$

So:

$$\lim_{n \rightarrow \infty} \frac{a^{n+1}}{1+a^{n+1}} : \frac{a^{n}}{1+a^{n}} =$$

$$\lim_{n \rightarrow \infty} \frac {a^{n+1}.(1+a^{n})}{a^{n}.(1+a^{n+1})} =$$

$$\lim_{n \rightarrow \infty} \frac {a.(1+a^{n})}{1+a.a^{n}} =$$

$$\lim_{n \rightarrow \infty} \frac {a}{a^{n+1} + 1} + \lim_{n \rightarrow \infty} \frac{a^{n+1}}{a^{n+1} + 1}$$

$$\lim_{n \rightarrow \infty} \frac {a}{a^{n+1} + 1} = \lim_{n \rightarrow \infty} \frac { \frac{a}{a}}{a^{n} + \frac{1}{a}} = \frac{1}{0 + \frac{1}{a}} = 0$$

$$\lim_{n \rightarrow \infty} \frac {a^{n+1}}{a^{n+1} + 1} = \lim_{n \rightarrow \infty} \frac {1}{1 + \frac{1}{a^{n+1}}} = 0$$

Thus

$$\lim_{n \rightarrow \infty} \frac {a^{n+1}}{a^{n+1} + 1} : \frac{a^{n}}{1+a^{n}} = 0 + 0 = 0$$

This way I proved the convergence for a in (-1, 1)

Is it ok?

Thank you for your comments.

 

[arildno]

$$\lim_{n \rightarrow \infty} \frac {a}{a^{n+1} + 1} = \lim_{n \rightarrow \infty} \frac { \frac{a}{a}}{a^{n} + \frac{1}{a}} = \frac{1}{0 + \frac{1}{a}} = 0$$

That limit is incorrect, the expression goes to "a", not to 0.
However, since a<1, convergence is achieved.

 

[twoflower]

That limit is incorrect, the expression goes to "a", not to 0.
However, since a<1, convergence is achieved.

You're right, I didn't take into account that a is constant.