Question about theta of $T(n)=4T(n/5)+n$ I have this recurrence relation $T(n)=4T(\frac{n}{5})+n$  with the base case $T(x)=1$  when $x\leq5$. I want to solve it and find it's $\theta$. I think i have solved it correctly but I can't get the theta because of this term $\frac{5}{5^{log_{4}n}}$ . Any help?
$T(n)=4(4T(\frac{n}{5^{2}})+\frac{n}{5})+n$
$=4^{2}(4T(\frac{n}{5^{3}})+\frac{n}{5^{2}})+4\frac{n}{5}+n$ 
$=...$ 
$=4^{k}T(\frac{n}{5^{k}})+4^{k-1}\frac{n}{5^{k-1}}+...+4\frac{n}{5}+n$ 
$=...$ 
$=4^{m}T(\frac{n}{5^{m}})+4^{m-1}\frac{n}{5^{m-1}}+...+4\frac{n}{5}+n$  Assuming $n=4^{m}$
$=4^{m}T(\lceil(\frac{4}{5})^{m}\rceil)+((\frac{4}{5})^{m-1}+...+1)n$
$=n+\frac{1-(\frac{4}{5})^{m}}{1-\frac{4}{5}}n=n+5n-n^{2}\frac{5}{5^{log_{4}n}}$ 
$=6n-n^{2}\frac{5}{5^{log_{4}n}}$  

An alternative approach is to prove that $T(n)\leqslant5n$ for every $n$. This holds for every $n\leqslant5$ and, if $T(n/5)\leqslant5(n/5)=n$, then $T(n)\leqslant4n+n=5n$. By induction, the claim holds.
On the other hand, $T(n)\geqslant n$ for every $n\gt5$, hence $T(n)=\Theta(n)$.
