I was able to prove this sum
$$\sum_{n=1}^\infty\frac{{1}}{n+3^n}$$
is convergent through the comparison test but I don't get how to find its sum.
2026-04-02 14:19:10.1775139550
Calculating $\sum\limits_{n=1}^\infty\frac{{1}}{n+3^n} $
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What we can obtain, also by hand, is a reasonable estimation for example by
$$\sum_{n=1}^\infty\frac{{1}}{n+3^n}\approx 0.392<\sum_{n=1}^3\frac{{1}}{n+3^n}+\sum_{n=4}^\infty\frac{{1}}{3^n}=$$$$=\frac14+\frac1{11}+\frac1{30}+\frac32-1-\frac13-\frac1{9}-\frac1{27}\approx 0.393$$