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My first thought was to use the root test, but it doesn't seem so, that it would be make the problem much more easier. The ratio test would be make the problem much more difficult, I even don't see any kind of ratio to use the comparison test.

All of these above I think that I should use the root test.

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If $f(x) = \frac{1+x}{2+x} $, $f(x) =1-\frac1{x+2} $.

If $-1 \le x \le 1$, $1 \le x+2 \le 3$ so $1 \ge \frac1{x+2} \ge \frac13$ or $0 \le 1-\frac1{x+2} \le \frac23 $.

Therefore the series converges since $2n-\log n \gt n$ so each term is less that $(2/3)^n$.


Added later:

To show that $n > \ln n$:

Let $h(x) = x-\ln(x)$. $h'(x) = 1-\frac1{x}$, so $h'(x) > 0$ for $x > 1$. Since $h(1) = 1$, $h(x) > 0$ for $x > 1$.

Actually, this proof works for $h(x) = x-\ln(x)-1$, so $x > \ln(x)+1$ for $x > 1$.

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  • $\begingroup$ very nice solution, but how can I easily prove that $2n-\log n \gt n$? $\endgroup$ – Zauberkerl Mar 21 '17 at 19:33
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    $\begingroup$ It's equivalent to $n > \log(n)$ which you can prove using calculus for example, but it's probably overkill. $\endgroup$ – mathreadler Mar 21 '17 at 19:43
  • $\begingroup$ @Zauberkerl: I added a proof. $\endgroup$ – marty cohen Mar 21 '17 at 21:52

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