$$\sum_{j=1}^n \frac{1}{j} $$ given in the question $n>0$ I've tried inputting numbers for n to get $1,\frac{1}{2},\frac{1}{3}......\frac{1}{n}$ but i dont know how to put this in an equation form i know that stuff exists like $\frac{n^2(n+1)^2}{4}$ for $k^2$ but nothing for $\frac{1}{k^2}$?
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See this wikipedia article. I don't think there is a nice expression. – Arthur May 12 '18 at 07:17
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1Possible duplicate of How to find the sum of this series : $1+\frac{1}{2}+ \frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}$ – SK19 May 12 '18 at 07:21
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1The formula you quote for the sum of $k^2$ in fact is for $k^3$. You will not find a similar simple form for the harmonic series. – Mark Bennet May 12 '18 at 07:22
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Let $H_n=1+\frac12+...+\frac1n$. There is no polynomial $f$ such that $f(n)=H_n$ for all $n$. One way of seeing this is that $H_n$ grows slower and slower as $n\to\infty$, but polynomials grow faster and faster as $n\to\infty$. Specifically, we have $H_{n+1}-H_n\to 0$, but if $f$ is a non-constant polynomial, then $f(n+1)-f(n)$ is also a polynomial, and either converges to plus or minus infinity if $\deg f\geq 2$, or is constant if $\deg f=1$.
Jack M
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