Hi all,(adsbygoogle = window.adsbygoogle || []).push({});

my task is to solve the convergence of the sum in dependance to the parameterareal.

[tex]

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

[/tex]

I did it this way:

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

Let's tryd'Alembert's criterion, which tells us this:

[tex]

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

[/tex]

So:

[tex]

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

[/tex]

[tex]

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

[/tex]

[tex]

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

[/tex]

[tex]

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

[/tex]

[tex]

\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

[/tex]

[tex]

\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

[/tex]

Thus

[tex]

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

[/tex]

This way I proved the convergence forain(-1, 1)

Is it ok?

Thank you for your comments.

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# Homework Help: Is this the right approach to solve it?

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