Find a value of n so that partial sum is greater equal to 10

Solution 1:

Using harmonic numbers $$S_{n} = \sum_{k=1}^n \frac{1}{k} =H_n$$ and you want to solve for $n$ the equation $$H_n=k$$ Using the asymptotics $$H_n=\gamma +\log (n)+\frac{1}{2 n}-\frac{1}{12n^2}+O\left(\frac{1}{n^4}\right)$$ you have as a very first approximation $$k=\gamma +\log (n)\quad \implies \quad n\sim e^{k-\gamma } \tag 1$$ Using one more term $$k=\gamma +\log (n)+\frac{1}{2 n}\quad \implies \quad n\sim -\frac{1}{2 W\left(-\frac{1}{2}e^{\gamma -k}\right)} \tag 2$$ where $W(.)$ is Lambert function.

For $k=10$, $(1)$ will give, as a real, $n_1=12366.968$ and $(2)$ will give $n_2=12366.468$ while the "exact" solution is $n=12366.470$.

Checking $$H_{12366}=9.999962148 \qquad \text{and} \qquad H_{12367}=10.00004301$$

I do not understand your friend's solution (which is obviously wrong).

Solution 2:

By condensation, I think the solution means the idea behind Cauchy's condensation test. Let $S_n = \displaystyle \sum_{k= 1} ^n f(k)$, where $f(k) = \dfrac 1k$. We notice $f$ is decreasing. Take a look in $S_n$ when $n$ is a power of 2 $$S_1 = f(1)$$ $$S_2 = f(1) + f(2)$$ $$S_4 = f(1) + f(2) + f(3) + f(4) > f(1) + f(2) + f(4) + f(4)$$ Here we changed $f(3)$ for $f(4)$, since $f(3) > f(4)$. $$S_8 = f(1) + f(2) + \cdots +f(8) > f(1) + f(2) + 2f(4) + 4f(8)$$ Here we changed $f(3)$ for $f(4)$ and $f(5),f(6), f(7)$ for $f(8)$. And so on. More generally we have:

$$S_{2^N} > f(1) + f(2) + 2f(4) + 4f(8) + 8f(16) + \cdots + 2^{N-1} f(2^N)$$

Since $f(k) = \dfrac 1k$ we have: $$f(1) + f(2) + 2f(4) + 4f(8) + 8f(16) + \cdots + 2^{N-1} f(2^N) = $$ $$= 1+ \dfrac 12 + \dfrac 24 + \dfrac 48 + \dfrac 8{16} + \cdots + \dfrac{2^{N-1}}{2^N}$$ $$ = 1 + \underbrace{ \dfrac 12 + \dfrac 12 + \cdots + \dfrac 12}_{N} = 1+\dfrac N2$$

Then $S_{2^N} > 1 + \dfrac N2$. Using $N = 18$ we get $S_{2^{18}} >1 + \dfrac{18}{2} = 10$.