Question

solve the 1/2 + 2/3 + 3/4 ... 99/100
i'm forget to solve it
hehehe

Answer 1

\[\sum_{n=1}^{99} \frac{ n }{ n+1 }\]

Answer 2

\[\sum_{n=1}^{99}[\frac{ n+1 }{ n+1 } - \frac{ 1 }{ n+1 }] =\sum_{n=1}^{99}[1 - \frac{ 1 }{ n+1 }]\]

Answer 3

does this help ?

Answer 4

n/n+1 is n - (1/n+1) so gets nearer and nearer to n in the limit.

Might as well just feed it into Wolfram.