Sequences and Series Question

**dwally89** · July 12th 2011, 12:15 PM

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 $(a_n)$ be given recursively by the formula:
$a_{n+1}=a_n+\frac{1}{n^2a_n}$ , where $a_1=1$

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

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

**Opalg** · July 12th 2011, 12:31 PM

Hint for (iii): induction (using (i), as you guessed).

**dwally89** · July 12th 2011, 01:24 PM

Originally Posted by Opalg

Hint for (iii): induction (using (i), as you guessed).

Is this correct?

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

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

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

which is clearly true.

**dwally89** · July 14th 2011, 05:05 AM

Originally Posted by dwally89

Is this correct?

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

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

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

which is clearly true.

Anyone?

**Archie Meade** · July 14th 2011, 06:06 AM

Originally Posted by dwally89

Is this correct?

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

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

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

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

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

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

then

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

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

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

which is clearly true.

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

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

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

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

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

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

If P(q) is true, then

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

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

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