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I need to take the limit of this summation so that I kind find out whether it converges or diverges. The equation is:

$$\sum_{k=1}^\infty \frac{4}{k+4}$$

What I have tried so far is the following:

$$\lim_{k \rightarrow \infty} \frac{k+4}{(k+1)+4}$$

Which then gives us $\frac{4}{4} = 1$, however I know that this is incorrect. How do your properly take the limit of the summation?

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5 Answers 5

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By the integral test: $\sum_{k=1}^{\infty} \frac{4}{k+4}>4\int_{1}^{\infty} \frac{1}{x+4}dx=\infty$.

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  • $\begingroup$ There is an easier way. $\endgroup$ May 3, 2015 at 2:34
  • $\begingroup$ This shows divergence directly without reference to the Harmonic series, though you are right in saying that the comparison test is more succinct =) $\endgroup$
    – Archaick
    May 3, 2015 at 2:37
  • $\begingroup$ I would say that because the series is a constant times a deleted harmonic series, the series diverges. I would say that is easier. $\endgroup$ May 3, 2015 at 2:38
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If you're just trying to find out weather it converges or diverges, you don't need to actually find its limit.

Instead try a comparison test with the Harmonic Series $$\sum_{n=1}^{\infty}\frac{1}{n}$$

This definitely diverges as it's a p-series with p=1

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Try this: $$\lim_{k\to \infty} \frac{4}{k+4} = \lim_{k\to \infty} \frac{\frac{4}{k}}{1+\frac{4}{k}} = 0$$ So the limit divergence test is inconclusive. But in general, that series diverges because it's a constant times a deleted series of a harmonic series.

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When n is large enough, \begin{align} \sum_{k=1}^n \frac{4}{k+4} &= 4\sum_{k=1}^n\frac{1}{k+4}\cr &= 4\sum_{k=5}^n\frac{1}{k}\cr \end{align}

Since we know $$\lim_{n \to \infty}\sum_{k=5}^n\frac{1}{k} = \infty$$

Then $\sum_{k=1}^\infty \frac{4}{k+4}$ diverges

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For $k\geq 1$, $$ \frac{4}{4+k}=\frac{1}{1+\frac{k}{4}}> \frac{1}{1+k} $$

But, for $N\geq 1$ $$ \sum_{k=1}^{N}\frac{1}{1+k}=\frac{1}{2}+\frac{1}{3} + \ldots + \frac{1}{N+1}= \big(\sum_{k=1}^{N}\frac{1}{k}\big)+ \big(\frac{1}{N+1}-1\big) $$

So, since the harmonic series diverges, given any $M>0$, you can find an index $N$ such that $\sum_{1}^{N} 1/k > M+1$, and since $1/(N+1)-1 > -1$, for that same index N, $\sum_{1}^{N}1/(1+k)>M$, and $\sum 1/(1+k)$ diverges.

Therefore, $\sum 4/(4+k)$ also diverges, by the comparison test.

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