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Find the solutions to this inquality by induction

  • Thread starter emyt
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Homework Statement


sum of i from i = 1 to n 1/i^2 <= 2




The Attempt at a Solution


I have a solution, that for any n the sum from i to n of 1/i^2 < sum 1/i(i-1) = 1/(i-1) - 1/i .. et c but this is not inductive.. can I get any hints? thank you
 

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  • #2
SammyS
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Homework Statement


[tex]\sum_{i=1}^{n}\ \frac{1}{i^2}\ \le\ 2[/tex]



The Attempt at a Solution


I have a solution, that for any n the sum from i to n of 1/i^2 < sum 1/i(i-1) = 1/(i-1) - 1/i ... etc but this is not inductive.. can I get any hints? thank you
Here is your original expression. Click on it to copy & edit your Original Post, if you like.

[tex]\sum_{i=1}^{n}\ \frac{1}{i^2}\le 2[/tex]

Certainly, 1 < 2, so it's true for the base case.

Assume that [tex]\sum_{i=1}^{n}\ \frac{1}{i^2}\le 2[/tex] is true for n≥1. Show that [tex]\sum_{i=1}^{n+1}\ \frac{1}{i^2}\le 2[/tex] is true.

You may also get some idea from looking at the sum for n=1, n=2, n=3, n=4, …

And/Or look at [tex]2\ -\ \sum_{i=1}^{n}\ \frac{1}{i^2}[/tex] for several values of n.
 
  • #3
SammyS
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...

The Attempt at a Solution


I have a solution, that for any n the sum from i to n of 1/i^2 < sum 1/i(i-1) = 1/(i-1) - 1/i .. etc but this is not inductive.. can I get any hints? thank you
I can't figure out what you you're saying here.

Certainly, [tex]\frac{1}{i^2}<\frac{1}{i(i-1)}\ \text{ for }\ i\ge2\ .[/tex]

[tex]\sum_{i=1}^{n}\ \frac{1}{i^2}=1+\sum_{i=2}^{n}\ \frac{1}{i^2}\ <\ 1+\sum_{i=2}^{n}\frac{1}{i(i-1)}=1+\sum_{j=1}^{n-1}\ \frac{1}{(j+1)j}[/tex]

As it turns out, [tex]\sum_{j=1}^{\infty}\ \frac{1}{(j+1)j}=1\ ,[/tex] so this should be do-able.
 
  • #4
SammyS
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Wolfram Alpha gives the following result:
[tex]\sum_{j=1}^{n-1}\ \frac{1}{(j+1)j}=1-\frac{1}{n}\ [/tex]

Proving that this is true is pretty straight forward using induction.

Note:
[tex]\sum_{j=1}^{(n+1)-1}\ \frac{1}{(j+1)j}=\frac{1}{(n+1)n}+\sum_{j=1}^{n-1}\ \frac{1}{(j+1)j}\ ,\text{ and }\frac{1}{(n+1)n}=\frac{n+1-n}{(n+1)n}=\frac{n+1}{(n+1)n}-\frac{n}{(n+1)n}[/tex]​
 

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