convergence of sequence

**altave86** · November 19th 2011, 07:57 PM

I am having troubles proving the following theorem. Suppose that a sequence of positive numbers satisfies for all n:

$0 \leq \frac{a_{n+1}}{a_n}\leq 1+\frac{1}{n^2}$

then this sequence converges.

I was given the hint

$1+x \leq e^x$

I tried plugging this hint, and tried to showed that If the sequence is monotone after some term, then the series converges since it is a bounded sequence. However, if this series continues to oscilate indefinitely, I haven't been able to show that it converges.

help please!

**chisigma** · November 20th 2011, 04:39 AM

Originally Posted by altave86

I am having troubles proving the following theorem. Suppose that a sequence of positive numbers satisfies for all n:

$0 \leq \frac{a_{n+1}}{a_n}\leq 1+\frac{1}{n^2}$

then this sequence converges.

I was given the hint

$1+x \leq e^x$

I tried plugging this hint, and tried to showed that If the sequence is monotone after some term, then the series converges since it is a bounded sequence. However, if this series continues to oscilate indefinitely, I haven't been able to show that it converges.

help please!

The sequence is the solution of the difference equation...

$a_{n+1}= \alpha_{n}\ a_{n}$ (1)

... that can be written as...

$\Delta_{n}= a_{n+1}-a_{n}= (\alpha_{n}-1)\ a_{n}= \beta_{n}\ a_{n}$ (2)

As illustrated in...

http://www.mathhelpforum.com/math-he...-i-188482.html

... the solution of (2) is...

$a_{n}= a_{1}+\sum_{k=1}^{n-1} \Delta_{k}$ (3)

In Your case is $0 \le \alpha_{n} \le 1+\frac{1}{n^{2}}$ and the solution is more clear if we examine two possible alternatives...

a) if we suppose that $1 \le \alpha_{n} \le 1+\frac{1}{n^{2}}$ , then applying (3) we obtain...

$\lim_{n \rightarrow \infty} |a_{n}| \le |a_{1}|\ \sum_{k=1}^{\infty} \frac{1}{k^{2}}$ (4)

... and the series in (4) converges...

b) if we suppose that $0<\alpha_{n} \le h <1$ , then is...

$\lim_{n \rightarrow \infty} a_{n} =0$ (5)

Kind regards

$\chi$ $\sigma$

**altave86** · November 20th 2011, 05:00 PM

THanks a lot! But I fail to see how the equence converges. In a), we know that if the limit exists, then the limit is not greater then the RHS of (4). However, the series might diverge in the interval [0,RHS(4)]

**chisigma** · November 21st 2011, 12:10 AM

Originally Posted by chisigma

The sequence is the solution of the difference equation...

$a_{n+1}= \alpha_{n}\ a_{n}$ (1)

... that can be written as...

$\Delta_{n}= a_{n+1}-a_{n}= (\alpha_{n}-1)\ a_{n}= \beta_{n}\ a_{n}$ (2)

As illustrated in...

http://www.mathhelpforum.com/math-he...-i-188482.html

... the solution of (2) is...

$a_{n}= a_{1}+\sum_{k=1}^{n-1} \Delta_{k}$ (3)

In Your case is $0 \le \alpha_{n} \le 1+\frac{1}{n^{2}}$ and the solution is more clear if we examine two possible alternatives...

a) if we suppose that $1 \le \alpha_{n} \le 1+\frac{1}{n^{2}}$ , then applying (3) we obtain...

$\lim_{n \rightarrow \infty} |a_{n}| \le |a_{1}|\ \sum_{k=1}^{\infty} \frac{1}{k^{2}}$ (4)

... and the series in (4) converges...

b) if we suppose that $0<\alpha_{n} \le h <1$ , then is...

$\lim_{n \rightarrow \infty} a_{n} =0$ (5)

A much better solution uses the concept of 'infinite product'. The solution of the difference equation...

$a_{n+1}= \alpha_{n}\ a_{n}$ (1)

... where $\alpha_{n}= \frac{a_{n+1}}{a_{n}}$ , can be written as...

$a_{n}= a_{1}\ \prod_{k=1}^{n-1} \alpha_{k}$ (2)

... so that the condition $0 \le \alpha_{n} \le 1+\frac{1}{n^{2}}$ leads us to the relation...

$\lim_{n \rightarrow \infty} |a_{n}| \le |a_{1}|\ \prod_{k=1}^{\infty} (1+\frac{1}{k^{2}})$ (3)

... and the 'infinite product' in (3) converges...

Kind regards

$\chi$ $\sigma$

**chisigma** · November 21st 2011, 12:16 AM

Originally Posted by altave86

THanks a lot! But I fail to see how the equence converges. In a), we know that if the limit exists, then the limit is not greater then the RHS of (4). However, the series might diverge in the interval [0,RHS(4)]

Better answer to Your question is given in my last post. Here the convergence of 'infinite sum' $\sum_{k=1}^{\infty} \frac{1}{k^{2}}$ means the convergence of 'infinite product' $\prod_{k=1}^{\infty} (1+\frac{1}{k^{2}})$ ...

Kind regards

$\chi$ $\sigma$

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