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I am looking specifically at the 'Test of Divergence' that states the following:

If $\lim_{x\to \infty} a_{n}$ does not exist or if $\lim_{x\to \infty} a_{n} \ne 0$, then the series $\sum_{n=1}^\infty a_{n}$ is divergent.

In the problem I am referring to, the given series is:

$$\sum_{k=1}^\infty\frac{n^7}{n^8 + 3}$$

It's not difficult to show that $$\lim_{x\to \infty} \frac{n^7}{n^8 + 3} = 0$$

So why does the given series above $diverge$? By test of divergence should it not converge?

Thanks.

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    $\begingroup$ Compare with $1/n$. Your test falls since it doesn't tell you that if the terms go to zero, the series converges, only that if they don't, the series doesn't. $\endgroup$ – Cameron Williams Aug 14 '19 at 3:19
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    $\begingroup$ The test given is an implication $P \rightarrow Q$. It is not a biconditional statement. So if $P$ is true then you can say $Q$ is true as well but if $P$ is false then the implication can be true regardless of the truth of $Q$. So if $\lim_{n \to \infty}a_n=0$, then you cannot infer anything about the convergence. $\endgroup$ – Anurag A Aug 14 '19 at 3:26
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The converse of the divergence test is false. I.e., there are divergent series $\Sigma a_n$ with $\lim_{n\rightarrow \infty} a_n=0$. The canonical example of this is the harmonic series $\Sigma \frac{1}{n}$.

We can show that your series diverges with the limit comparison test. Given series $\Sigma a_n$ and $\Sigma b_n$ with nonnegative terms, the limit comparison test states that if $0<\lim_{n\rightarrow \infty} \frac{a_n}{b_n}<\infty$, then either both series converge or both series diverge.

Observe the following: $$\lim_{n\rightarrow\infty} \frac{\frac{n^7}{n^8+3}}{\frac{1}{n}} = \lim_{n\rightarrow\infty} \frac{n^8}{n^8+3} = 1<\infty$$ Then your series diverges because $\Sigma \frac{1}{n}$ diverges.

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Apply Integral test for convergence and the result is immediate:

$\displaystyle \int_1^\infty \frac{x^7~ dx}{x^8 + 3} \to \infty$

i.e., the integral diverges.

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The test of divergence specifically states that $$\textrm{if } \sum_{n=1}^{\infty} a_n \textrm{ converges, then } \lim_{n\to \infty} a_n =0$$ which is the same as saying that $$\textrm{if } \lim_{n\to \infty} a_n \neq 0 \textrm{ then } \sum_{n=1}^{\infty} a_n \textrm{ diverges}$$ The implications are important, in general, that $a_n \to 0$ as $n \to \infty$ does not always mean that the series must to converges (e.g. consider the harmonic series).

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Test of divergence is not what you think of. The limit for n to infinity of your series is 0, but it doesn’t say that it converges, another example is the harmonic series 1+1/2+1/3+1/4+..., similarly, the limit is 0, but it diverges.

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$$ \frac{{n^7 }} {{n^8 + 3}} \geqslant \frac{{n^7 }} {{n^8 + n^8 }} = \frac{1} {{2n}} $$ as log as $$n\geq 2$$. By comparison test with harmonic series the given series is divergent. The condition $$ \mathop {\lim }\limits_{n \to \infty } a_n = 0 $$ is a necessary but NOT a sufficient condition.

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