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I have a lot of sum questions right now ... could someone give me the convergence of, and/or formula for, $\sum_{n=2}^{\infty} \frac{1}{n^k}$ when $k$ is a fixed integer greater than or equal to 2? Thanks!!

P.S. If there's a good way to google or look up answers to these kinds of simple questions ... I'd love to know it...

Edit: Can I solve it by integrating $\frac{1}{x^k}$ ? I can show it converges, but to find the formula? Is my question just the Riemann Zeta function?


Thanks guys! This got me the following result:

$\sum_{p} \frac{1}{p} \log \frac{p}{p-1} ~ ~ \leq ~ \zeta(2)$

summing over all primes $p$. (And RHS is Riemann zeta function.)

First, sum over all integers $p$ instead of primes. Then transform the log into $\sum_{m=1}^{\infty} \frac{1}{m p^{m}}$ (reference: wikipedia. I know.). Now we have (with rearranging):

$\leq ~ \sum_{m=1}^{\infty} \frac{1}{m} \sum_{p=2}^{\infty} \frac{1}{p^{m+1}}$

By the result of this question (Arturo's answer), this inner sum, which is $\zeta(m+1)$ is at most $\frac{1}{m+1-1} = \frac{1}{m}$. So we have

$\leq ~ ~ \sum_{m=1}^{\infty} \frac{1}{m} \frac{1}{m} = \zeta(2)$

I think this is a very pretty little proof. Thanks again, hope you math people enjoyed reading this....

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Try an integral test. You might also be interested in the Basel problem (don't miss Robin Chapman's notes linked to right at the bottom of the page) and the Riemann zeta function. – t.b. Apr 19 '11 at 1:34
joriki's answer is useful. – J. M. Apr 19 '11 at 1:57
up vote 2 down vote accepted

The fact that these series converge follows from the integral test; their values are somewhat more complicated, as indicated by the comments: they correspond to evaluating Riemann's zeta function at $k$, and this is highly nontrivial, even for positive integral $k$ (called the zeta constants). There are formulas for even integer values of $k$, though in terms of Bernoulli numbers. Note that your series is actually one less than the zeta constants, since $$\zeta(k) = \sum_{n=1}^{\infty}\frac{1}{n^k}$$ and your sum starts with $n=2$, not $n=1$.

As mentioned, the Integral Test shows these series converge. But to do it explicitly:

Consider the sequence of partial sums $$s_n = \sum_{r=2}^n \frac{1}{r^k}.$$ Since all summands are positive, this is an increasing sequence: $$s_2\leq s_3\leq s_4\leq\cdots$$ so the convergence of the sequence of partial sums is equivalent to showing that the sequence is bounded.

To show the sequence is bounded, consider the function $\frac{1}{x^k}$. This function is decreasing, so if you approximate $$\int_1^b\frac{1}{x^k}\,dx$$ using a right hand sum, you will get an underestimate for the integral.

Consider $$\int_1^{n}\frac{1}{x^k}\,dx,$$ and approximate it using a right hand sum with the interval $[1,n]$ divided into $n-1$ equal parts (so we break up the integral into $[1,2]$, $[2,3],\ldots,[n-1,n]$). The right hand sum with that partition is given by: $$\mathrm{Right Sum} = \frac{1}{2^k} + \frac{1}{3^k} + \cdots + \frac{1}{n^k} = s_{n}.$$ Since right hand sums are underestimates, then for every positive integer $n$ we have that: $$s_n \leq \int_1^n\frac{1}{x^k}\,dx = \frac{1-n^{1-k}}{k-1}.$$ Since the bounds get larger for larger $n$, we have: $$s_n\leq \frac{1-n^{1-k}}{k-1} \leq \lim_{n\to\infty}\frac{1 - n^{1-k}}{k-1} = \frac{1}{k-1}.$$ So the sequence of $s_n$ is bounded above. Since it is increasing, the sequence of partial sums converges.

Since the sequence of partial sums converges, the series converges as well.

(This is the essence of the Integral Test: if you have a series $$\sum a_n$$ such that there is a positive, continuous, decreasing function $f(x)$ such that $f(n)=a_n$ for all $n$, then the convergence of the series is equivalent to the convergence of the improper integral $$\int_1^{\infty}f(x)\,dx$$ in the sense that if the integral converges, then the series converges; and if the integral diverges, then the series diverges. Of course, you can change the lower limit of the integral if necessary.)

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Wow, thanks for the thought-out answer. To clarify, this also proves that the sequence converges to something less than $\frac{1}{k-1}$, right? Thanks again! – usul Apr 19 '11 at 2:04
@b01024: The argument shows the limit of the series is at most $\frac{1}{k-1}$, but not necessarily something strictly less than $\frac{1}{k-1}$. In fact, the limit is striclty smaller, but the argument above does not show it. – Arturo Magidin Apr 19 '11 at 2:09
Thanks, and thank you for the precision in answering (I was sloppy). – usul Apr 19 '11 at 2:11
If you are interested, this allowed me to show that $\sum_{p} \frac{1}{p} \log \frac{p}{p-1}$ , I'll edit my question to show the steps! – usul Apr 19 '11 at 2:15

Yes, this sum converges, and yes, you can prove this using the integral test. For $k$ even the exact value (plus $1$) turns out to be a rational multiple of $\pi^k$ (see, for example, Wikipedia), but for odd $k$ the exact values are much more mysterious.

The best way to find answers to these simple questions is to learn the basic techniques for solving them. In the context of discussing the convergence of series this means learning the basic convergence tests, as well as the knowledge of several specific examples such as might be covered in a typical calculus course. Beyond that, questions like this are hard to google for, so your best option is to ask them here!

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Thank you. I'm in the unfortunate position of having forgotten most of these techniques just when I need them again, so this is helping me pick it back up! Thanks for the answer! – usul Apr 19 '11 at 2:42

You are looking for the Riemann zeta function $\zeta(k)$ (or close to it: the sum usually starts at $n=1$).

Since you are supposed to be doing research for a class project, perhaps you should search for it.

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Gotcha. That's really interesting because my problem is related to counting the primes.... Thanks! – usul Apr 19 '11 at 1:52

It's also easy to show that this series converges through more elementary means (though without getting a good estimate of the value), by a version of the same elementary technique often used to show that the harmonic series diverges. Start by fixing $k=2$; since the terms for any other $k$ are smaller than this, then clearly if this series converges the others will. Next, break it into chunks of length $2^n$: $S = {1\over 1^2} + ({1\over 2^2}+{1\over 3^2}) + ({1\over 4^2}+{1\over 5^2}+{1\over 6^2}+{1\over 7^2}) + \ldots$ Now, replace each of the terms in the sets of parentheses by the first term there; since we know that e.g. ${1\over 7^2} \lt {1\over 4^2}$, our sum will be bounded by the result; $S\lt S'$, where $S' = {1\over 1^2} + ({1\over 2^2}+{1\over 2^2}) + ({1\over 4^2}+{1\over 4^2}+{1\over 4^2}+{1\over 4^2}) + \ldots$ - and here it's easy to see that, for instance, the $4$ terms of $1\over 4^2$ will add up to $1\over 4$, the (unshown) $8$ terms of $1\over 8^2$ will add up to $1\over 8$, etc; so $S'$ is just $1+{1\over 2}+{1\over 4}+{1\over 8}+\ldots = 2$, and so we know that the original series $S$ converges (and in fact that its value is less than $2$).

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The easiest way to prove that $1+\frac{1}{2^2}+...+\frac{1}{n^2}$ is convergent, is proving by induction that

$$ 1+\frac{1}{2^2}+...+\frac{1}{n^2} \leq 2 -\frac{1}{n} \,.$$

If $k >2$, convergence follows immediately from $\frac{1}{j^k} \leq \frac{1}{j^2}$.

I always find fascinating the fact that the above inequality is a trivial induction problem, while the easier inequality: $ 1+\frac{1}{2^2}+...+\frac{1}{n^2} \leq 2 \,.$ cannot be proven by induction....

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This is one interesting example where proving more is easier. But I prefer the less mysterious argument, where we upper bound $\frac{1}{n^2}$ by $\frac{1}{n(n-1)}$ and sum the latter series easily by telescoping. Is this what you mean by "proving by induction"? – Srivatsan Jul 26 '11 at 11:22

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