The test I know are Cauchy's root test, Cauchy's integral, Raabe's test, logarithmic test and D' Alembert's Ratio Test. I dont know which test I can use to prove that this series converges? $$\sum \left( (n^3+1)^{\frac{1}{3}} -n \right) $$
Asked
Active
Viewed 95 times
-1
-
1I wanted to mark your question as a duplicate of this one, but it has no accepted answers, so I couldn't. – PinkyWay Jul 01 '20 at 20:26
-
1Furthermore, Find x for which $\sum [(n^3+1)^\frac{1}{3}-n]x^n$ converges.. – PinkyWay Jul 01 '20 at 20:32
-
Hint: use the mean value theorem on $f(x)=(n^3+x)^{1/3}$ on the interval $x\in[0,1]$. – Greg Martin Jul 01 '20 at 20:33
-
1Convergence for the series: hint – PinkyWay Jul 01 '20 at 20:34
-
1Also, this closed question – PinkyWay Jul 01 '20 at 20:36
1 Answers
3
If we can prove that $$ (n^3+1)^\frac13-n \leq \frac1{n^2}, $$ then we can use the comparison test to prove that the series converges.
Let's find $x$ such that $$ (n^3+x)^\frac13-n \leq \frac1{n^2}. $$ $$ \begin{align} (n^3+x)^\frac13 &\leq \frac1{n^2}+n\\ n^3+x &\leq \left(\frac1{n^2}+n\right)^3\\ n^3+x &\leq n^3+3+\frac{3}{n^{3}}+\frac{1}{n^{6}}\\ x &\leq 3+\frac{3}{n^{3}}+\frac{1}{n^{6}}\\ & \leq 3. \end{align} $$ Since in your particular case $x=1\leq3$, we know that the sum is less than the sum of $1/n^2$, so it must converge.
Polygon
- 1,854
-
There already exist five answered posts dealing with the same queation. – PinkyWay Jul 01 '20 at 20:37
-
-
-
@IPPK I missed that. I just saw a few more obscure and complicated tests mentioned without reading carefully, so I assumed that OP knew the more simple and intuitive tests. – Polygon Jul 01 '20 at 20:43
-
Heyy sorry i didn't see a duplicate of this question and I checked their answer out, but none made it as simple and straightforward as this one. Thank you so much seriously god bless, been looking everywhere for a simple answer – RiRi Jul 01 '20 at 20:47