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I am currently writing my IB HL Math Internal Assessment. I was researching why "$\sum_{n=1}^\infty \frac{1}{n}$" diverges , but "$\sum_{n=1}^\infty \frac{1}{n^2}$" converges. My research brought me to the p-Series convergence test: $$\sum_{n=1}^\infty \frac{1}{n^x}$$ which converges when x>1.

The convergence of this type of infinite series does not only depend on IF the values of the sequence converge to zero. The convergence of this type of infinite series actually depends on the RATE AT WHICH the values of the sequence converge to zero (Why does the harmonic series diverge but the p-harmonic series converge). I want to further explore the correlation between the rate at which the sequence converges to zero and the convergence of the series. For example, I tried to create a function to examine the trend of the the rates at which $\sum_{n=1}^\infty \frac{1} {n^x}$ converge to zero, but realized that the rates of the graphs are not one value (i.e. 4).. Each rate of $\frac{1}{n^x}$ are infinite values, such as $y=\frac{-0.5}{n^2}$ (the rate of convergence of the sequence of the harmonic series).

The problem I am having is that I cannot find research online that tells me more than "the convergence of the series "1/n^x" is dependent on the rate at which the values converge to 0. I want to further explore the "rigorous proofs" that were mentioned in "Why does the harmonic series diverge but the p-harmonic series converge".

I tried to explain this topic as best I could. Thank you for your time.

EDIT: the SEQUENCE of 1/n would be: 1, 1/2, 1/3, etc... The SERIES of 1/n would be 1 + (1/2) + (1/3) + ... + (1/n) Here is an example of what I am looking for: The terms of the harmonic series (1/n) converge to zero, but the summation of all these terms, called the series, DOES NOT converge to a number. This is because the sequence does not converge to zero fast enough. I am trying to gain a deeper mathematical understanding of this idea.

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Y. Moondhra
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  • I think this is a really good question, so +1. – Simply Beautiful Art Jan 15 '17 at 01:55
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    To be clear: you seem to be asking for more than just a proof that $\sum_{n=1}^\infty \frac{1}{n^p}$ converges for $p > 1$, you seem to be asking about how fast that happens? – Chris Jan 15 '17 at 01:58
  • @Chris No, I get out of it that the OP wants even more specifically the answer to the question in general, as in, beyond just that series. I think that would make this very interesting. – Simply Beautiful Art Jan 15 '17 at 02:01
  • Why not just compare with the integral of $1/t^p$? Assuming $p>0$, you can draw a picture to see that $$\sum_{n=1}^{\infty} \frac{1}{n^p} \geq \int_1^{\infty} \frac{1}{t^p}dt \geq \sum_{n=2}^{\infty} \frac{1}{n^p}$$ – Michael Jan 15 '17 at 02:12
  • @Michael It does not tell you how fast it converges though, and how "fast-ness" relates to convergence. – Simply Beautiful Art Jan 15 '17 at 02:13
  • @SimpleArt : Wouldn't how fast it converges depend on the tail of the integral? You do not have to start the above integral inequality at $n=1$, you can start at $n=m$. – Michael Jan 15 '17 at 02:14
  • @Michael I think the OP's question is more about the rate the summand goes to 0. – Simply Beautiful Art Jan 15 '17 at 02:15
  • @SimpleArt : I don't get what you are saying. The summand is $1/n^p$ so the summand goes to zero like that. – Michael Jan 15 '17 at 02:17
  • @Michael But just because $a_n\to0$ does not mean $\sum a_n$ converges. OP wants to know how fast $a_n\to0$ means the sum will converge. – Simply Beautiful Art Jan 15 '17 at 02:18

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