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Determine the convergence of $$\sum_{n=1}^{\infty} (\sqrt[n]{2n^2}-1)^n$$

I have no idea what test I should use any hint? I think is divergent, I was thinking using comparison test but don't know where to start. But for fun I got:

$(\sqrt[n]{2n^2}-1)^n$ ~ 1 for large n. therefore, the series should be divergent. so I need to show that $a_n \gt b_n$ for some $b_n$ diverges. for all $n \in N$, we have $(\sqrt[n]{2n^2}-1)^n$ $\gt$ n. since $\sum_{n=1}^{\infty}$ $n$ is divergent then $\sum_{n=1}^{\infty}$ $(\sqrt[n]{2n^2}-1)^n$ is divergent

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You should also have the test (insert name) $$ \lim_{n \to \infty} a_n \not= 0 $$ implies that $\sum a_n$ does not converge. So you just need to show this is true.

EDIT:

Let $b_n = (2n^2)^{1/n} - 1$. If we can show that $\exists \, N$ where $b_n < 1$ for all $n > N$, then we have shown the sum converges.

Claim 1: $\lim_{n \to \infty} b_n = 0$. (Show this.)

Claim 2: $\exists N$ such that $b_n < 1$ for all $n \ge N$ (this follows from Claim 1. Why?)

We can then split the sum into $$ \sum_{n=1}^{N-1} b_n^n + \sum_{n=N}^{\infty} b_n^n $$ The first term is finite (denote it $L_1$), the second term, the $b_n < 1$. Let $\delta < 1$ be bound on these $b_n$ then we can bound the second term by $$ \sum_{n=N}^{\infty} b_n^n \le \sum_{n=N}^{\infty} \delta^n = L_2 $$ Which is a geometric series (hence finite). This proves the sequence converges.

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  • $\begingroup$ little bite more hint...so i cant use $\sqrt[n]{2n^2}-1)^n$ $\gt$ n? or $\sqrt[n]{2n^2}-1)^n$ ~1 $\endgroup$ – jenny Dec 5 '14 at 3:31
  • $\begingroup$ You absolutely could if you could show that those inequalities were true (note: this would imply the divergent test, since both of those limits would not converge to 0) $\endgroup$ – Gregory Dec 5 '14 at 3:33
  • $\begingroup$ sorry i was trying to use comparison test, but i can use divergent test..so how would i show that is inequalities? $\endgroup$ – jenny Dec 5 '14 at 3:38
  • $\begingroup$ like this.. there exists $n_o$ $\in N$ s.t $\vert \sum_{n=1}^{\infty} \vert (\sqrt[n]{2n^2}-1)^n$ $\lt$ $\epsilon$ for all m,n $\in$ N with m $\gt$n $\gt$$n_o$? $\endgroup$ – jenny Dec 5 '14 at 3:42
  • $\begingroup$ I will edit the answer to be more helpful. $\endgroup$ – Gregory Dec 5 '14 at 3:49

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