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Is \(\displaystyle \sum_{n=1}^\infty\frac{1}{n}\) possible? Its a harmonic progression and so there is no direct formula for its sum. I have seen some sites using zeta function to find its sum but zeta function works only if the exponent of n is greater than 1,right?

Note by Shubham Srivastava
4 years, 2 months ago

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The harmonic series is divergent and has been proven to be so for at least several hundred years. Proving this should be easy for any competent mathematics student.

One of my favorite proofs uses a simple algebraic inequality, which is valid for \( n > 1 \): \[ \frac{1}{n-1} + \frac{1}{n+1} = \frac{2n}{n^2-1} > \frac{2n}{n^2} = \frac{2}{n}. \] Consequently, \[ \frac{1}{n-1} + \frac{1}{n} + \frac{1}{n+1} > \frac{3}{n}, \] for all \( n > 1 \). Now suppose \[ H_m = \sum_{n=1}^m \frac{1}{n}, \] for \( m \ge 1 \). Then using the inequality we just proved, \[ \begin{align*} H_{3m+1} &= \sum_{n=1}^{3m+1} \frac{1}{n} \\ &= 1 + \sum_{n=1}^m \left( \frac{1}{3n-1} + \frac{1}{3n} + \frac{1}{3n+1} \right) \\ &> 1 + \sum_{n=1}^m \frac{1}{n} \\ &= 1 + H_m. \end{align*} \] In particular, this implies \( H_4 > 2 \), \( H_{13} > 3 \), \( H_{40} > 4 \), and in general, \[ H_{(3^k-1)/2} \ge k \] for all positive integers \( k \). This directly proves \( \displaystyle\lim_{m \to \infty} H_m \) is unbounded, because for any \( k > 1 \), we can find \( m(k) = (3^k-1)/2 \) such that \( H_{m(k)} > k \). Therefore the given infinite series is divergent.

What I like about this proof is that it furnishes a surprisingly tight lower bound on the partial sums without relying on calculus; namely, \( H_m > \log_3 (1+2m) \). No doubt, this is much worse than the bound \( H_m > \gamma + \log m \), but a proof of the latter involves significantly heavier mathematical machinery than what we have used here. There exist numerous other similar and/or simpler proofs (e.g. Oresme's well-known groupings by successive powers of 2), but this one is a bit more obscure and so is more interesting to me.

The zeta function has a pole of order 1 at \( s = 1 \), and whose residue is 1. This is the unique singularity of \( \zeta(s) \) and it is not removable. There's no purpose in trying to use the zeta function to evaluate the harmonic series, because the divergence of the series is precisely why the zeta function behaves the way it does. Hero P. · 4 years, 2 months ago

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@Hero P. Thanks!!! Shubham Srivastava · 4 years, 2 months ago

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There is another well known proof that the harmonic series is divergent: we do this by chopping increasingly large, but always finite 'chunks' with sum at least 1/2 as follows:

C_0 = {1}

C_1 = {2}

C_2 = {3,4}

C_3 = {5,6,7,8}

...

C_i = {2^(i-1), ... 2^(i)}

Observe that the sum of reciprocals in every set C_i is at least 1/2; and for any bound B, we can take the first 2B sets; the sum of reciprocals will then exceed this bound B. This bound is logarithmic, but probably inferior to Hero's

Edit: Hero has mentioned this proof in his response (Oresme); the bound is indeed inferior. Using his definition of H_m, we have:

Hm > (1/2) log2 (m) = log_4 (m)

log4 (m) < log3(m) < log_3 (1 + 2m) Gabriel Wong · 4 years, 2 months ago

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Here's yet another proof that \(\sum_{n\ge 1} \frac 1 n\) is divergent, using integration.

Imagine rectangles \([1, 2]\times [0, 1], [2, 3] \times[0, \frac 1 2], \ldots [n-1, n]\times [0, \frac 1 {n-1}]\). These rectangles totally cover the graph of \(y = \frac 1 x\) from x=1 to x=n. Hence the area, \(1 + \frac 1 2 + \frac 1 3 + \ldots + \frac 1{n-1}\) is greater than \(\int_1^n \frac 1 x\cdot dx = \log n\) which diverges. C Lim · 4 years, 2 months ago

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