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Finding Convergence of series $$\sum^{\infty}_{k=1}\frac{3k}{k^2+4}$$ using Integral test or Divergences Test.

What i try

Let $$\sum^{\infty}_{k=1}\frac{3k}{k^2+4}=3/5+6/8+\sum^{\infty}_{k=3}\frac{3k}{k^2+4}$$

Let $\displaystyle f(x)=\frac{3x}{x^2+4}.$ Then $f'(x)<0$ for $x>2$

So using Integral Test

$$\int^{\infty}_{3}\frac{3x}{x^2+4}dx=1.5\ln|x^2+4|\bigg|^{\infty}_{3}\rightarrow -\infty$$

So series is Diverges.

But when i apply Divergence Test.

In $$\sum^{\infty}_{k=1}\frac{3k}{k^2+4}$$. Then its limit goes to $0$

Means this series is converges.

Plese tell me which one is right. Thanks

jacky
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    Your second argument is not valid. What is the statement of the Divergence Test? – Kavi Rama Murthy May 06 '20 at 08:28
  • Means first one is Right. Professor – jacky May 06 '20 at 08:29
  • The correct wording is "this series converges" or "is convergent". –  May 06 '20 at 08:30
  • Remember, it's literally a test for divergence. It tells you nothing about convergence. Only a non zero limit is informative (meaning divergence). A zero limit gives no information. – Deepak May 06 '20 at 08:31
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    @jacky You wrote $\to -\infty$ instead of $\to \infty$. Otherwise your first method is correct. Also, there is a much simpler method to show divergence. – Kavi Rama Murthy May 06 '20 at 08:34

3 Answers3

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Here is a way you can verify the series diverges .

$\dfrac{3k}{k^2+4}\gt\dfrac{3k}{(k+2)^2}=\dfrac{k+2k}{(k+2)^2}\gt\dfrac{k+2}{(k+2)^2}=\dfrac{1}{k+2}$

$\implies\displaystyle\sum\dfrac{3k}{k^2+4}\gt\displaystyle\sum\dfrac1k=\infty$

Jimmy
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The divergence test is conclusive when the limit of the sequence is nonzero. When this limit is zero, there is nothing you can say.

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$$\sum^{\infty}_{k=1}\frac{3k}{k^2+4}> \sum^{\infty}_{k=1}\frac{3k}{k^2}=3\sum^{\infty}_{k=1}\frac 1k$$

As the harmonic series, $\sum_{k=1}^\infty \frac1k$, is divergent, it follows that the series diverges.

Axion004
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