Induction

Induction

Postby Guest » Mon Apr 09, 2018 1:22 pm

Use the fact that [tex]\frac{1}{k}[/tex] - \frac{1}{k+1} = \frac{1}{k(k+1)} to show that

n
sigma (\frac{1}{k(k+1)} = 1- \frac{1}{n+1}
r=1

How do I solve this?

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Re: Induction

Postby nathi123 » Mon Apr 09, 2018 3:27 pm

[tex]\frac{1}{k}-\frac{1}{k+1}=\frac{k+1}{k(k+1)}-\frac{k}{k(k+1)} =\frac{k+1-k}{k(k+1} =\frac{1}{k(k+1)}[/tex] (1)
[tex]\Rightarrow A= \sum_{k=1}^{n }\frac{1}{k(k+1)}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n(n+1)}[/tex]
From (1) [tex]\Rightarrow A= \frac{1}{1}-\frac{1}{2}+\frac{1}{2} - \frac{1}{3} +\frac{1}{3} - \frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}[/tex]
[tex]\Rightarrow A=\sum_{k=1}^{n }\frac{1}{k(k+1)}= 1 - \frac{1}{n+1} = \frac{n}{n+1}[/tex]. :)

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Re: Induction

Postby Guest » Tue Apr 10, 2018 8:24 am

nathi123 wrote:[tex]\frac{1}{k}-\frac{1}{k+1}=\frac{k+1}{k(k+1)}-\frac{k}{k(k+1)} =\frac{k+1-k}{k(k+1} =\frac{1}{k(k+1)}[/tex] (1)
[tex]\Rightarrow A= \sum_{k=1}^{n }\frac{1}{k(k+1)}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n(n+1)}[/tex]
From (1) [tex]\Rightarrow A= \frac{1}{1}-\frac{1}{2}+\frac{1}{2} - \frac{1}{3} +\frac{1}{3} - \frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}[/tex]
[tex]\Rightarrow A=\sum_{k=1}^{n }\frac{1}{k(k+1)}= 1 - \frac{1}{n+1} = \frac{n}{n+1}[/tex]. :)



The induction step is hidden in the ...s. The key idea (using (1) to telescope the sum) that makes the induction work is of course correct, but if your instruction was explicitly to use induction, you'll want to replace the ... handwaving with a formal induction proof like

[tex]\frac{1}{1*2} = 1-\frac{1}{2}[/tex] so it's valid when n = 1.

Suppose that for some [tex]l\ge2[/tex],

[tex]\sum_{k=1}^{l-1 }\frac{1}{k(k+1)}= 1 - \frac{1}{l}[/tex].

Then [tex]\sum_{k=1}^{l}\frac{1}{k(k+1)}= \left(1 - \frac{1}{l}\right)
+ \frac{1}{l(l+1)}[/tex] (add the next term, corresponding to [tex]k=l[/tex])

[tex]= 1 - \frac{1}{l} + \left(\frac{1}{l} - \frac{1}{l+1}\right)[/tex] (Simplify the sum using the induction hypothesis and apply (1) to the last term)

[tex]= 1-\frac{1}{l+1}[/tex]. (Regroup and cancel the [tex]1/l[/tex] terms) Showing that the same identity holds when one additional term is included in the sum.

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Re: Induction

Postby nathi123 » Tue Apr 10, 2018 11:47 am

The way I have demonstrated proves the affirmation without the use of mathematical induction.
I do not think it is appropriate to demonstrate this method using (1).
I suggest the following formulation. To prove that
[tex]\sum_{k=1}^{n }\frac{1}{k(k+1)}=\frac{n}{n+1}[/tex] using mathematical induction.

If n=1[tex]\Rightarrow\frac{1}{1.2}=\frac{1}{2} -[/tex]true
Let's assume that [tex]=\frac{n}{n+1}[/tex] (2)and we will prove that
[tex]\sum_{k=1}^{n +1}\frac{1}{k(k+1)}=\frac{n+1}{n+2}[/tex]
[tex]\sum_{k=1}^{n +1}\frac{1}{k(k+1)}=\sum_{k=1}^{n }\frac{1}{k(k+1)} + \frac{1}{(n+1)(n+2)}=\frac{n}{n+1}+\frac{1}{(n+1)(n+2)}=\frac{n^{2}+2n+1}{(n+1)(n+2)}[/tex]
[tex]\Rightarrow \sum_{k=1}^{n +1}\frac{1}{k(k+1)}=\frac{(n+1)^{2}}{(n+1)(n+2)}=\frac{n+1}{n+2}[/tex].

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