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This is probably incredibly simple, but we've just started the topic, and we've just gone over geometric series, p-series, and harmonic series. Its simple when the series is given explicitly in sigma-notation, but I struggle when they don't give you the form and just give you the first few numbers. The question exactly is:

Determine whether the following series converges or diverges. Give a reason for your answer.

$$1+\frac15+\frac19+\frac1{13}\dots$$

Any tips/hints/help would be much appreciated.

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    $\begingroup$ I think your self-diagnosis is accurate: you want to master the skill of being given a series in this form and writing it as a series given by a formula. How would you describe, in words, the pattern of denominators in this series you're given? $\endgroup$ – Greg Martin Feb 4 '17 at 6:29
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    $\begingroup$ Thanks for the reply. The first thing that I noticed was that it was all very close to multiples of 5, but off by an increasing amount. (1/9 is close to 1/(10-1), 1/13 is 1/(15-2)...) For whatever reason, I didn't see that it was also one larger than multiples of 4, which is a lot easier to write explicitly as tilper showed in the answer below. I guess as I practice, I'll recognize these things more easily? $\endgroup$ – akot717 Feb 4 '17 at 6:36
  • $\begingroup$ Absolutely—the more we practice, the better we get, just like anything! $\endgroup$ – Greg Martin Feb 4 '17 at 8:02
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The series you give looks like: $$\sum_{n=0}^{+\infty} \frac1{4n+1}$$

To see this, note that each of the denominators are $1$ larger than a multiple of $4$: $5 = 4 + 1$, and $9 = 8 + 1$, and $13 = 12 + 1$. Presumably this pattern continues. Then we note that $4 = 4 \cdot 1$, and $8 = 4 \cdot 2$, and $12 = 4 \cdot 3$, and so on. So the first few terms are really:

$$1 + \frac1{4 \cdot 1 + 1} + \frac1{4 \cdot 2 + 1} + \frac1{4 \cdot 3 + 1} + \cdots$$

Being able to identify this type of pattern is a skill that will come naturally with practice. So I definitely recommend getting lots of practice!

Anyway, you can use the Limit Comparison Test with the harmonic series to get the answer to the actual question. Let me know if you require further guidance.


EDIT: The way you originally thought about it can also work: \begin{align*} 1+\frac15+\frac19+\frac1{13}+\cdots &= 1+\frac1{5-0}+\frac1{10-1}+\frac1{15-2}+\cdots\\[0.3cm]&=1+\frac1{5\cdot 1-0}+\frac1{5\cdot2-1}+\frac1{5\cdot3-2}+\cdots\\[0.3cm]&=\sum_{n=0}^{+\infty}\frac1{5\cdot n-(n-1)}\\[0.3cm]&=\sum_{n=0}^{+\infty}\frac1{4n+1}\end{align*}

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  • $\begingroup$ Thanks for the answer tilper. For whatever reason, instead of recognizing that each term was one larger than a multiple of four, I saw this pattern: 1/9 is 1/(10-1), 1/13 is 1/(15-2), 1/17 is 1/(20-3). Interesting. I suppose as I practice, I will think simpler first, and that way I'll avoid skipping over simple solutions. Once you have $\sum_{n=0}^{+\infty} \frac1{4n+1}$ I can see that it is in the form $\sum_{n=0}^{+\infty} \frac1{an+b}$ and this behaves like a harmonic series (diverging to infinity). Right? $\endgroup$ – akot717 Feb 4 '17 at 6:41
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    $\begingroup$ @akot717, yes, that's exactly right. In fact, "behaves like" is the exact term I use when I teach/tutor this type of thing in person. I can see you're building an intuition for this, and that's great. Also, re: the original way you were thinking of it - that was a bit more complicated but it can be made to work also. See my edit to the post. $\endgroup$ – tilper Feb 4 '17 at 15:19
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    $\begingroup$ thanks for all the help and encouragement! Also thanks for the edit. All makes sense now. Cheers $\endgroup$ – akot717 Feb 5 '17 at 8:43
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For all $n$, we have $$\frac{1}{4n+\color{red}1}\geq \frac{1}{4n+\color{red}n}=\frac{1}{5}\cdot\frac{1}{n}.$$ Thus, using the Comparison Test, the series $$\frac{1}{5}+\frac{1}{9}+\frac{1}{13}+\cdots$$ is divergent since the series $\sum_{n=1}^{+\infty}(\frac{1}{5}\cdot\frac{1}{n})$ is divergent. Since adding a finite number of terms in a divergent series does not affect divergence, the given series $$1+\frac{1}{5}+\frac{1}{9}+\frac{1}{13}+\cdots$$ is also divergent.

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  • $\begingroup$ This is what I was thinking, even when I didn't recognize that it was $\frac1{4n+1}$. I thought, this is close to $\frac1{5n}$, and that is clearly divergent, so this should be divergent too since the sum of the corresponding terms are greater than the terms in $\frac1{5n}$. $\endgroup$ – akot717 Feb 4 '17 at 6:44
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You could also observe that

$$1 +\frac{1}{5}+\frac{1}{9}\cdots \geq 1 + \frac{1}{5}\left (1 +\frac{1}{2}+\frac{1}{3} +\cdots\right)$$ The term inside brackets is the harmonic series which....

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The series can be written as : $$\sum_{n=0}^\infty\frac{1}{4n+1}$$

The integral test is the simplest in this case. The series is a lot like the harmonic series, so one might think that integral test can be used.

Compute the integral: $$\int\frac{1}{4x+1}dx=\frac{\ln(4x+1)}{4}+C$$

Then compute the improper integral: $$\int_1^\infty\frac{1}{4x+1}dx=\lim_{n\to\infty}\frac{\ln(4n+1)}{4}-\frac{\ln5}{4}$$ Clearly, the integral is divergent. Hence, the series is also divergent.

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  • $\begingroup$ We haven't learned the integral test, and I'm limited to explaining in terms of geometric, p-series, or harmonic series properties. But thanks for the explanation! $\endgroup$ – akot717 Feb 4 '17 at 6:43

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