Estimate following sums as the functions of variable $n$:
a) $\displaystyle\sum_{i=1}^{n}e^i\ln i$
b) $\displaystyle\sum_{i=1}^{2n}(-1)^i\ln i$
c) $\displaystyle\sum_{i\ge 0}^{}\frac{\ln(n+i)}{e^i}$
d) $\displaystyle\sum_{i\ge 0}^{}\frac{(-1)^i}{\ln(n+i)}$
I prefer the easiest ways to do such things, so Euler-Maclaurin formula isn't in my style. Are these examples very hard? I want to learn how to do such things but unfortunately when it comes to solving I have no ideas.
Only in b) I thought I knew but the idea is probably useless: $\sum_{i=1}^{2n}(-1)^i\ln i=\sum_{i=1}^{n}\ln(2i)-\ln(2i-1)$ so let: $f(x)=\ln(2x)-\ln(2x-1)$ and it's easy to check that $f(x)$ non increasing for $x\ge 1$ then: $\int_i^{i+1}f(x)\le f(i)\le\int_{i-1}^if(x)$ it follows: $\int_2^{n+1}f(x)\le\sum_{i=2}^n f(i)\le\int_1^{n}f(x)$ and integral is easy to count: $\int f(x)=\frac{1}{2}\left( 2x\ln(2x)-(2x-1)\ln(2x-1)-1 \right)+C$ But is it worth anything? Can I write from this $\sum_{i=1}^n f(i)$ using asymptotics notations like big $O$ or $\Theta$?
I don't know how to approach the rest.