# Sequences and Series Question

• Jul 12th 2011, 12:15 PM
dwally89
Sequences and Series Question
Hi,

I am new to this forum so please let me know if I am posting in the wrong place, and I will move my thread.

I have the following question:
Let the sequence $\displaystyle (a_n)$ be given recursively by the formula:
$\displaystyle a_{n+1}=a_n+\frac{1}{n^2a_n}$ , where $\displaystyle a_1=1$

i) Prove that $\displaystyle a_n\ge1$ for all natural numbers $\displaystyle n$
ii) Prove that $\displaystyle a_{n+1}\ge{a_n}$ for all natural numbers $\displaystyle n$
iii) Prove that $\displaystyle a_n\le1+\sum_{k=1}^{n-1} \frac{1}{k^2}$ for all $\displaystyle n\ge2$

I have managed to do parts i) and ii), but am unable to do part iii)
I think part i) is meant to help somehow.

Help would be appreciated,

Thanks
• Jul 12th 2011, 12:31 PM
Opalg
Re: Sequences and Series Question
Hint for (iii): induction (using (i), as you guessed).
• Jul 12th 2011, 01:24 PM
dwally89
Re: Sequences and Series Question
Quote:

Originally Posted by Opalg
Hint for (iii): induction (using (i), as you guessed).

Is this correct?

Proof by induction
Prove for $\displaystyle n=2$
$\displaystyle a_2=a_{1+1}=a_1+\frac{1}{n^2a_1}=1+1=2$
$\displaystyle 1+\sum_{k=1}^1 \frac{1}{k^2}=1+1=2$
Case holds for $\displaystyle n=2$

Assume case holds for $\displaystyle n=q$, i.e.
$\displaystyle a_q\le1+\sum_{k=1}^{q-1} \frac{1}{k^2}$

Prove for $\displaystyle n=q+1$
$\displaystyle a_{q+1}=a_q+\frac{1}{n^2a_q}$
$\displaystyle a_{q+1}\le 1+\sum_{k=1}^{q-1} \frac{1}{k^2}+\frac{1}{n^2a_q}$
$\displaystyle a_{q+1}\le a_{q+1}+\sum_{k=1}^{q-1} \frac{1}{k^2}+\frac{1}{n^2}$
$\displaystyle 0\le \sum_{k=1}^{q-1} \frac{1}{k^2}+\frac{1}{n^2}$

which is clearly true.
• Jul 14th 2011, 05:05 AM
dwally89
Re: Sequences and Series Question
Quote:

Originally Posted by dwally89
Is this correct?

Proof by induction
Prove for $\displaystyle n=2$
$\displaystyle a_2=a_{1+1}=a_1+\frac{1}{n^2a_1}=1+1=2$
$\displaystyle 1+\sum_{k=1}^1 \frac{1}{k^2}=1+1=2$
Case holds for $\displaystyle n=2$

Assume case holds for $\displaystyle n=q$, i.e.
$\displaystyle a_q\le1+\sum_{k=1}^{q-1} \frac{1}{k^2}$

Prove for $\displaystyle n=q+1$
$\displaystyle a_{q+1}=a_q+\frac{1}{n^2a_q}$
$\displaystyle a_{q+1}\le 1+\sum_{k=1}^{q-1} \frac{1}{k^2}+\frac{1}{n^2a_q}$
$\displaystyle a_{q+1}\le a_{q+1}+\sum_{k=1}^{q-1} \frac{1}{k^2}+\frac{1}{n^2}$
$\displaystyle 0\le \sum_{k=1}^{q-1} \frac{1}{k^2}+\frac{1}{n^2}$

which is clearly true.

Anyone?
• Jul 14th 2011, 06:06 AM
Re: Sequences and Series Question
Quote:

Originally Posted by dwally89
Is this correct?

Proof by induction
Prove for $\displaystyle n=2$
$\displaystyle a_2=a_{1+1}=a_1+\frac{1}{n^2a_1}=1+1=2$
$\displaystyle 1+\sum_{k=1}^1 \frac{1}{k^2}=1+1=2$
Case holds for $\displaystyle n=2$

Assume case holds for $\displaystyle n=q$, i.e.
$\displaystyle a_q\le1+\sum_{k=1}^{q-1} \frac{1}{k^2}$

Prove for $\displaystyle n=q+1$
$\displaystyle a_{q+1}=a_q+\frac{1}{n^2a_q}$

******************************************

What you should be doing from here is showing that "if"

$\displaystyle a_q\le 1+\sum_{k=1}^{q-1}\frac{1}{k^2}$

then

$\displaystyle a_{q+1}\le 1+\sum_{k=1}^{q}\frac{1}{k^2}$

******************************************

$\displaystyle a_{q+1}\le 1+\sum_{k=1}^{q-1} \frac{1}{k^2}+\frac{1}{n^2a_q}$
$\displaystyle a_{q+1}\le a_{q+1}+\sum_{k=1}^{q-1} \frac{1}{k^2}+\frac{1}{n^2}$
$\displaystyle 0\le \sum_{k=1}^{q-1} \frac{1}{k^2}+\frac{1}{n^2}$

which is clearly true.

$\displaystyle P(q);\;\;\;\;a_q\le 1+\sum_{k=1}^{q-1}\frac{1}{k^2}$

$\displaystyle P(q+1);\;\;\;\;a_{q+1}\le 1+\sum_{k=1}^{q}\frac{1}{k^2}$

Show that "if" P(q) is true, "then" P(q+1) must be true.
Write P(q+1) in terms of P(q).

$\displaystyle a_{q+1}=a_q+\frac{1}{q^2a_q}$

$\displaystyle a_q+\frac{1}{q^2a_q}\le\ 1+\sum_{k=1}^{q}\frac{1}{k^2}\;\;?$

$\displaystyle a_q+\frac{1}{q^2a_q}\le\ 1+\sum_{k=1}^{q-1}\frac{1}{k^2}+\frac{1}{q^2}\;\;?$

If P(q) is true, then

$\displaystyle a_q+\frac{1}{q^2a_q}\le\ a_q+\frac{1}{q^2}\;\;?$

Now, you may use the result to part (i) to justify the truth of this.