How do you solve $\displaystyle \frac{n}{n+1}$ questions?

Question

$1/2 + 2/3 + 3/4 + 4/5... + 99/100.$

I've checked a lot of websites, but I still quite can't understand how do you do it. I think you're supposed to convert everything into a common denominator?

Please help, I really appreciate your answers.

Edit: After much thinking and discussions with my teachers and my friends, I came up with this equation:

1/2 + 2/3 + 3/4 + 4/5 +... + 99/100 = (1- 1/2)+(1- 1/3)+(1- 1/4)+...+(1- 1/100) = 99-(1/2 + 1/3 + 1/4 +... + 1/100) = 99-(1/(2+3+4+...+100) = 99-(1/5049) = 98+(5048/5049)

However, when I try to find the sum, I end up with 94.8126, like what Claude did. However, without using a calculator, how can I get the right answer as a fraction (mixed number)? There must be something wrong with my equation and I can't find it. Please help.

Answer 1

To summarize the discussion in the comments:

This sum will not have a convenient closed form. This is because it is essentially a Harmonic Number for which closed forms are not known.

Specifically, your sum is $$\sum_{n=1}^{99}\frac n{n+1}=\sum_{n=1}^{99}\left(1-\frac 1{n+1}\right)=\sum_{n=1}^{99}1-\sum_{n=1}^{99}\frac 1{n+1}=99-\sum_{n=1}^{99} \frac 1{n+1}$$ and the final summation is a Harmonic number. (well, technically, it is $H_{100}-1$, but it is essentially a Harmonic number).

Answer 2

Interesting could be the generalization $$S_p=\sum_{n=1}^{p}\frac n{n+1}=p+1-H_{p+1}$$ Now, if $p$ is large, using the asymptotics of harmonic numbers will give $$S_p=p+(1-\gamma) -\log \left({p}\right)-\frac{3}{2 p}+\frac{13}{12 p^2}+O\left(\frac{1}{p^3}\right)$$ Computed exactly $$S_{99}=\frac{264414864639329557497913717698145082779489}{2788815009188499086581352357412492142272}= 94.8126225$$ while the above truncated expansion would give $$S_{99}\sim \frac{11759431}{117612}-\gamma -\log (99)=94.8126235$$