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I am trying to determine if the series $\frac{1}{5} + \frac{1}{8} + \frac{1}{11} + \frac{1}{14} + \frac{1}{17} $ is convergent or divergent?

I can see that the terms in the denominator differ by 3 each time.

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marked as duplicate by Dietrich Burde, Fly by Night, Aqua, Clarinetist, samjoe Nov 24 '17 at 18:32

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    $\begingroup$ Compare to $\frac{1}{3}\sum_{k=2}^{\infty} \frac{1}{n}$. $\endgroup$ – user328442 Nov 24 '17 at 16:09
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    $\begingroup$ Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. $\endgroup$ – Fly by Night Nov 24 '17 at 16:10
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If we include the term $\frac12$ at the beginning, then each term is larger than $\frac1{3n}$. Thus any partial sum is larger than $\frac13H_n$, where $H_n=\sum_{i=1}^n\frac 1i$ are the harmonic numbers.

The harmonic numbers are known to diverge to infinity, so the sequence $\frac 13H_n-\frac12$ must also diverge. By the comparison test, your series diverges.

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Note that the terms are greater than the terms of $$\frac 16+\frac 19+\frac 1{12}+\frac 1{15}+\cdots =\frac 13\left(\frac 12+\frac 13+\frac 14+\frac 15+\cdots\right)$$

When we have terms of the form $\cfrac 1{an+b}$, choose a fixed $r$ with $ar\gt b$ and consider terms of the form $\cfrac 1{a(n+r)}$

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  • $\begingroup$ @MichaelHardy Fair comment. I started putting just the first part and ended up writing more than I intended. $\endgroup$ – Mark Bennet Nov 24 '17 at 18:39
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The pattern for the denominator is found by first adding another term so that the series starts $\frac{1}{2} + \frac{1}{5} + \frac{1}{8} + ...$ then for the series $2 + 5 + 8 + 11 ... $ the general nth term is found as $ T_n = a + (n-1)d$ where $a = \frac{1}{2}$ is the first term and $d = 3$ the difference. So $T_n = 3n -1$ and the original series is found to be $\sum _{n=1}^\infty \frac{1}{3n-1}$

If we let $f(x) = \frac{1}{3x-1}$ then $f'(x) = \frac{-1}{(3x-1)^2} $ which is negative for all $n \geq1$, which means it is decreasing on $[1,\infty)$. This function is also positive for all $n \geq1$ and so the conditions are met for the integral test.

$$\int_1^\infty f(x) =\lim _{b\to\infty}\int_1^b \frac{1}{(3x-1)} $$ $$= \lim _{b\to\infty}\frac{1}{3}\ln(3x-1)|_1^b$$ $$= \lim _{b\to\infty}\frac{1}{3}[\ln(3b-1) - \ln(2)]$$ $$=\infty$$

By the integral test the series $\sum _{n=1}^\infty \frac{1}{3n-1}$ is therefore divergent.

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    $\begingroup$ I think you mean divergent...? $\endgroup$ – abiessu Nov 24 '17 at 16:07
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    $\begingroup$ The series diverges. $\endgroup$ – user328442 Nov 24 '17 at 16:12

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