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I've been having a lot of trouble with this sequence

$$a = {4 \sqrt[n]{n^3} + \frac{ 10^{5n} }{ n! } }$$

I first tried the Nth term test but I didn't even know how to start. I was looking ahead in the chapters and found that

$$ \lim_{n\to \infty} \sqrt[n]{n} = 1$$ and $$ \lim_{n\to \infty} \frac{x^n }{ n!} = 0$$

But I found these rules in series and I wonder if I can use them with sequences.

Any help would be appreciated thank you.

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  • $\begingroup$ The rules that you state relate to sequences, there are no series here. $\endgroup$ – Yves Daoust Oct 14 '18 at 19:36
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Yes of course you can use the same way you use for the series, notably we can prove

Notably

$$b_n=c_n^{\frac1n}=\sqrt[n]{n^3}\to 1$$

since

$$\frac{c_{n+1}}{c_n}=\frac{(n+1)^3}{n^3}\to 1$$

and

$$d_n= \frac{ 10^{5n} }{ n! } \to 0$$

indeed

$$\frac{d_{n+1}}{d_n}= \frac{ 10^{5n+5} }{ (n+1)! } \frac{n!} {10^{5n} }=\frac{10^5}{n+1}\to 0$$

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  • $\begingroup$ So could I say that the limit of the sequence equals 1, so by the Nth term test it diverges? $\endgroup$ – Kode Ch Mar 4 '18 at 16:49
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    $\begingroup$ You can simply show that limit of sequence is 1 by the criteria I've indicated, do you know them, Of course from here you can conlude that tje series $\sum a_n$ diverges since the necessay condition $a_n \to 0$ does not hold. $\endgroup$ – gimusi Mar 4 '18 at 16:52
  • $\begingroup$ This is nice!!:)) $\endgroup$ – Peter Szilas Mar 6 '18 at 8:40
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Using the result that the sum of convergent sequences is convergent, we need only deal with the first term of $\{a_n\}$ since $\dfrac{ 10^{5n} }{ n! } $ converges to $0$ from your second identity.

Now write $4\sqrt[n]{n^3}=4(n^{1/n})^3$ and prove that it converges. This proves the convergence of $\{a_n\}$.

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