Evaluating $T(n)=T(n-1)+c\log(n)$

Question

$$T(n)=T(n-1)+c\log(n)$$ where $T(1)=d$

$$T(n)-T(n-1)=c\log(n)$$ $$T(n)-T(1)=c\sum_{i=1}^{n}\log(n)$$ $$T(n)-d=c\cdot n\cdot \log(n)$$ $$T(n)=c\cdot n\cdot \log(n)+d$$

So $T(n)=O(n\log(n))$

Is it correct?

Answer 1

It seems to be $$ T(n) =c \log(n! e^{d}). $$

Answer 2

The equation $$ T(n)=T(n-1)+c\cdot\ln(n) $$ yields $$ T(n-1)=T(n-2)+c\cdot\ln(n-1)\\ T(n-2)=T(n-3)+c\cdot\ln(n-2)\\ \vdots $$ We can conclude $$ T(n)=T(n-1)+c\cdot\ln(n)=T(n-2)+c\cdot\ln(n-1)+c\cdot\ln(n)=\ldots\\ =T(1)+c\sum_{k=2}^n\ln(k) =d+c\sum_{k=2}^n\ln(k). $$

If we simplify $$ \sum_{k=2}^n\ln(k)=\ln(n!), $$ then we get $$ T(n)=d+c\cdot\ln(n!). $$