Why isn't there a proper solution on internet on this sum:
$$\sum_{i=1}^{n} \frac{1}{i}$$
Can you help me with your solution?
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Simply put, no closed formula is known. Please see Is there a partial sum formula for the Harmonic Series? – WaveX Mar 20 '19 at 01:44
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1Look up Harmonic Numbers. The relation to the logarithm function is discussed in the answers to Showing that $\lim_{n\to\infty}\sum^n_{k=1}\frac{1}{k}-\ln(n)=0.5772\ldots$ – robjohn Mar 20 '19 at 01:44
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But isn't there a proper formula for every value of n considering that every "relative" of this equation has it? – Adnan C Mar 20 '19 at 01:49
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What do you mean by "relative"? – robjohn Mar 20 '19 at 01:50
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I mean that there is a formula for every sum with alternation in the denominator from this one? – Adnan C Mar 20 '19 at 01:52
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Not really. If you sum the reciprocalds of all positive squares you get $\pi^2/6$, but what if you take only a finite number of terms? The partial sum as a function of the numver of terms is a complicated thing that the simple expression for the limiting value, $\pi^2/6$, conceals. – Oscar Lanzi Mar 20 '19 at 01:52
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But this sum looks really simple for finding its formula for solution with n in it. – Adnan C Mar 20 '19 at 01:59
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Also, the sum to infinity of $\sum\limits_{k=1}^n\frac1k$ diverges; that is, it grows without bound, getting closer to $\log(n)+\gamma$ where $\gamma$ is the Euler Mascheroni Constant. – robjohn Mar 20 '19 at 02:00
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I mean there should be a simple formula for finding every nth harmonic number right? – Adnan C Mar 20 '19 at 02:01
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1There are approximations that get pretty close. One is $$\sum_{k=1}^n\frac1k\sim\log(n)+\gamma+\frac1{2n}-\frac1{12k^2}+\frac1{120k^4}$$ – robjohn Mar 20 '19 at 02:02
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2If you can find a "simple" formula you will be famous among mathematicians. – DanielWainfleet Mar 20 '19 at 03:25