Proving the Summation of an Infinite Series

[andyk23]

1. Homework Statement

∑ i=1 to n1+(1/i2)+(1/(1+i)2)−−−−−−−−−−−−−−−−−−−−√ = n(n+2)/n+1

2. The attempt at a solution

First I did the base case of p(1) showing 3/2 on the LHS equals the 3/2 on the RHS.
Then I assumed p(k) and wrote out the formula with k in it.
Then prove p(k+1)= p(k)+ 1+1/(k+1)2+1/(k+2)2−−−−−−−−−−−−−−−−−−−−−−√
=k(k+2)/k+1 + 1+1/(k+1)2+1/(k+2)2−−−−−−−−−−−−−−−−−−−−−−√
Then I squared each to get rid of the square root.
(k(k+2)/(k+1))^2+ (k+1)^2/(k+1)^2 + 1/(k+1)^2 + 1/(k+2)^2
Now I'm stuck any Guidance would be great thanks!

 

[SammyS]

1. Homework Statement

∑ i=1 to n1+(1/i2)+(1/(1+i)2)−−−−−−−−−−−−−−−−−−−−√ = n(n+2)/n+1

The above line is nonsensical. Not enough parentheses, probably missing exponentiation, dangling radical ...

Do you mean $\displaystyle \sum_{i=1}^{n}\sqrt{1+1/(i^2)+1/(1+i)^2}\ \ = \ \ n(n+2)/(n+1)\ ?$

If so, then I must be a much better mind reader than I have ever been given credit for !

2. The attempt at a solution

First I did the base case of p(1) showing 3/2 on the LHS equals the 3/2 on the RHS.
Then I assumed p(k) and wrote out the formula with k in it.
Then prove p(k+1)= p(k)+ 1+1/(k+1)2+1/(k+2)2−−−−−−−−−−−−−−−−−−−−−−√
=k(k+2)/k+1 + 1+1/(k+1)2+1/(k+2)2−−−−−−−−−−−−−−−−−−−−−−√
Then I squared each to get rid of the square root.
(k(k+2)/(k+1))^2+ (k+1)^2/(k+1)^2 + 1/(k+1)^2 + 1/(k+2)^2
Now I'm stuck any Guidance would be great thanks!

 

[andyk23]

yes sorry about that!

 

[SammyS]

yes sorry about that!

You answered "yes" to what? Please be precise .

 

[andyk23]

Do you mean $\displaystyle \sum_{i=1}^{n}\sqrt{1+1/(i^2)+1/(1+i)^2}\ \ = \ \ n(n+2)/(n+1)\ ?$

If so, then I must be a much better mind reader than I have ever been given credit for !

Yes I meant that!

 

[SammyS]

As you stated: For the induction step, you assume that

$\displaystyle \sum_{i=1}^{k}\sqrt{1+\frac{1}{i^2}+\frac{1}{(1+i)^2}}\ = \ \frac{k(k+2)}{k+1}$​

for k ≥ 1 . From that assumption, show that it follows that

$\displaystyle \sum_{i=1}^{k+1}\sqrt{1+\frac{1}{i^2}+\frac{1}{(1+i)^2}}\ = \ \frac{(k+1)(k+3)}{k+2}\ .$​

So look at $\displaystyle \ \ \sum_{i=1}^{k+1}\sqrt{1+\frac{1}{i^2}+\frac{1}{(1+i)^2}}\,,\ \$ which you can write as $\displaystyle \ \ \sqrt{1+\frac{1}{(k+1)^2}+\frac{1}{(k+2)^2}}+\sum_{i=1}^{k}\sqrt{1+\frac{1}{i^2}+\frac{1}{(1+i)^2}}\ .$

Now use you assumption, then do some algebra.

 

[andyk23]

I'm just having a trouble showing, $\frac{k(k+2)}{k+1} + \sqrt{1+\frac{1}{(k+1)^2}+\frac{1}{(k+2)^2}}\ .$ = $\frac{k+1(k+1)}{k+1+2}\ .$

 

[andyk23]

I'm just having a trouble showing, $\frac{k(k+2)}{k+1} + \sqrt{1+\frac{1}{(k+1)^2}+\frac{1}{(k+2)^2}}+\sum_{i=1}^{k}\sqrt{1+\frac{1}{i^2}+\frac{1}{(1+i)^2}}\ .$ = [itex]\frac{k+1(k+1+2)}{k+1}\ .[itex][/itex][/itex]

 

[andyk23]

I'm just having a trouble showing, $\frac{k(k+2)}{k+1} + \sqrt{1+\frac{1}{(k+1)^2}+\frac{1}{(k+2)^2}}\$ = $\frac{k+1(k+1+2)}{k+1}\$

 

[Michael Redei]

What you really need to show next is
$$
\frac{k(k+2)}{k+1}+\sqrt{1+\frac1{(k+1)^2}+\frac1{(k+2)^2}} = \frac{\color{red}{(k+1)} (k+3)}{k+\color{red}2}.
$$
I'd try subtracting $\frac{k(k+2)}{k+1}$ on both sides and then squaring the results.