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Using camparasion or limit camparasion test

Finding whether the series $$\sum^{\infty}_{k=1}\frac{5k^9-k^5+8\sqrt{k}}{3k^{11}-k^4+2}$$ is converge or diverge.

What i try

$$\sum^{\infty}_{k=1}\frac{5k^9-k^5+8\sqrt{k}}{3k^{11}-k^4+2}<\sum^{\infty}_{k=1}\frac{5k^9-k^5+8\sqrt{k}}{3k^{11}-k^4}<\sum^{\infty}_{k=1}\frac{5k^9-k^5+8\sqrt{k}}{3k^{11}+3k^{11}}$$

So here individual series is Diverge.

So our original series is diverge.

Is my process is right. If not then please tell How i solve it. Thanks

jacky
  • 5,194

1 Answers1

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  1. "comparison"
  2. Increasing the denominator decreases the fraction, so your final inequality is backwards: $$\sum^{\infty}_{k=1}\frac{5k^9-k^5+8\sqrt{k}}{3k^{11}-k^4}\not < \sum^{\infty}_{k=1}\frac{5k^9-k^5+8\sqrt{k}}{3k^{11}+3k^{11}}$$
  3. What makes you think $$\sum^{\infty}_{k=1}\frac{5k^9-k^5+8\sqrt{k}}{3k^{11}+3k^{11}}$$ diverges? You better look hard at this.
  4. Even if you were right about the previous two items, knowing for two series $A$ and $B$ that $A \le B$ and that $B$ diverges does not indicate that $A$ diverges. (You also apparently have the comparison test backwards.)

The first rule for using comparison tests is that you need to compare to a series whose behavior you know. Comparing to another series whose behavior is unknown and then just declaring it diverges proves nothing. Nobody will be fooled by your declaration.

The limit comparison test will be simpler for this particular example. I suggest you start off by multiplying top and bottom of the original fraction by $\frac 1{k^{11}}$, then consider what is going to happen as $k \to \infty$. Which terms dominate for high $k$? The other terms can be dropped from your comparison series.

Paul Sinclair
  • 43,643