0
$\begingroup$

$\frac{1}{3}+\frac{2}{6}+\frac{3}{11}+\frac{4}{18}+\frac{5}{27}+\dots$

I tried representing this in a sequence from and This is what I ended up with :

$$a_{n+1}=\frac{n}{2n+1-a_n}$$

But couldn't end up anywhere

  • 1
    Hint: $\sum 2k+1 = n^2$ – 2017-02-19
  • 2
    @Exodd Huh? I don't see how that helps? – 2017-02-19
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    Your formula for the general term $a_{n+1}=\frac{n}{2n+1-a_n}$ is wrong – 2017-02-19

2 Answers 2

3

Hint
The sequence is $a_n=\frac{n}{n^2+2}$, try to compare with $\frac{1}{n+2}$

0

Hint: you can also write $a_n=\frac{n}{n^2+2}$