Simple summation How can I resolve this summation : 
$$\sum_{k=0}^{a}\frac{1}{k+10}$$
I know I can simplify like this :
$$\sum_{k=0}^{a}\frac{1}{k+10}=\sum_{k=10}^{a+10}\frac{1}{k}$$
But I don't know what to do after...
 A: Exactly is the same spirit as Joel's answer $$S=\sum_{k=b}^a \frac{1}{k+c} = H_{a+c}-H_{b+c-1}$$ As you could see here, the harmonic number can be approximated by $$H_n=\gamma +\log(n) +\frac{1}{2 n}-\frac{1}{12
   n^2}+\cdots$$ Applied to the case $b=0$, $c=10$, this would give
$$S  \approx \frac{1}{2 (a+10)}+\frac{1}{12 (a+10)^2}+\log
   \left(\frac{a+10}9\right)-\frac{53}{972}$$ Let us try using $a=20$; the exact value is $\approx 1.166019$ while the approximation gives $1.166205$.
Let us try using $a=200$; the exact value is $\approx 3.097734$ while the approximation gives $3.097739$
A: There is not much insofar as a closed formula for this sort of summation. We write $H(N)=\sum_{n=1}^N \frac1n$ and call $H(N)$ the $N$-th hamonic number.
In your case we have $$\sum_{k=0}^a \frac{1}{k+10} = H(a+10)-H(9).$$
For large values of $a$, an approximation of the form $$H(N) \approx 1+ \int_1^N \frac{1}{x} dx = 1+\ln(N)$$ may be helpful. This provides an upper bound of $H(N)$. If the 1 is not added, this is a lower bound for $H(N)$.
Also $H(9) = 7129/2520$.
Thus we may approximate this quantity as $$\sum_{k=0}^a \frac{1}{k+10} \approx 1+\ln(a+10) - \frac{7129}{2520}.$$
