Mathematical Induction, Want to check I'm getting this right

Question

I just want to make sure i'm doing this right for other questions. i have a question: $$\sum\limits_{j=1}^{n}\frac{1}{j(j+1)} = \frac{1}{n(n+1)}$$ and have to show that the function $$f(n) = \frac{n}{n+1}$$ applies for all n in N. so it's the same setup right? where i do the n =1. then assume that it's true for n and prove that its true for n + 1 by simply saying $$\frac{n+1}{n+2}$$ right?

Answer 1

First, show that this is true for $n=1$:

$\sum\limits_{k=1}^{1}\frac{1}{k(k+1)}=\frac{1}{1+1}$

Second, assume that this is true for $n$:

$\sum\limits_{k=1}^{n}\frac{1}{k(k+1)}=\frac{n}{n+1}$

Third, prove that this is true for $n+1$:

$\sum\limits_{k=1}^{n+1}\frac{1}{k(k+1)}=$

$\color\red{\sum\limits_{k=1}^{n}\frac{1}{k(k+1)}}+\frac{1}{(n+1)(n+1+1)}=$

$\color\red{\frac{n}{n+1}}+\frac{1}{(n+1)(n+1+1)}=$

$\frac{n+1}{n+1+1}$

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Please note that the assumption is used only in the part marked red.