<s>[INST] Convergence of a sequence $\frac{1}{1+n^3}$ How can I prove by integral test that the sequence $1+ \dfrac{1}{1+2^3} + \dfrac{1}{1+3^3} + \dots + \dfrac{1}{1+n^3}$ is convergent?
Thank you. Is there a way that I can integrate $\dfrac{1}{1+n^3}$ ?
 [/INST] I will guess that the $n$-th term of the sequence is
$$\frac{1}{1+1^3}+\frac{1}{1+2^3}+\frac{1}{1+3^3}+\cdots+\frac{1}{1+n^3}.$$
There is indeed a closed form for $\int \frac{1}{1+t^3}\,dt$. However, the details are somewhat unpleasant. It is simpler to note that the $n$-th term is less than $\frac {1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+\cdots +\frac{1}{n^3}$
and use the fact that $\int_1^\infty \frac{1}{t^3}\,dt$ converges. 
Alternately, we can use the fact that the $n$-th term is less than or equal to 
$$\frac{1}{1+1^2}+\frac{1}{1+2^2}+\frac{1}{1+3^2}+\cdots+\frac{1}{1+n^2},$$
and use the fact that $\int_0^\infty \frac{1}{1+t^2}\,dt$ converges. 
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