Finding the Sum of series $S = \sum_{n=1}^{243} \frac{1}{n^{4/5}} $. If $S = \sum_{n=1}^{243} \frac{1}{n^{4/5}} $. 
Find the value of $\lfloor S \rfloor$ where $\lfloor \cdot \rfloor$ represents the greatest integer function.
By approximation using definite integral, I know the answer lies in the set $[10,15]$ (approximately) but I don't know how to find the exact sum.
 A: You're getting a very loose approximation with the integral if you start at $0$, because $\int_0^1 x^{-4/5}\,dx = 5$. That's the majority of the error.
To avoid this, write $$S = 1 + \sum_{n=2}^{243} \frac1{n^{4/5}}$$ and then use integrals to put upper and lower bounds on the sum from $2$ to $243$. Since the function $f(x) = \frac1{x^{4/5}}$ is decreasing for $x>0$, we have $$\int_2^{244} \frac1{x^{4/5}}\,dx < \sum_{n=2}^{243} \frac1{n^{4/5}} < \int_1^{243} \frac1{x^{4/5}}\,dx$$ and that should be enough to estimate the sum closely enough to find the floor $\lfloor S\rfloor$.
A: You will not find an exact value. You are probably expected to find an approximation. 
You could use the trapezium rule to approximate the integral 
$$\int_1^{243}f(x)dx \approx  \frac12(f_1+f_2)+\frac12(f_2+f_3)+...\frac12(f_{241}+f_{242})+\frac12(f_{242}+f_{243})$$ $$= -\frac12(f_1+f_{243})+\sum_1^{243}f_n $$
Then
$$I=\int_1^{243}x^{-4/5}dx=[5x^{1/5}]_1^{243}=5(3-1)=10$$ 
Therefore
$$\sum_1^{243}\frac{1}{n^{4/5}} \approx I+\frac12(\frac{1}{1^{4/5}}+\frac{1}{243^{4/5}})=10+\frac12(1+\frac{1}{81}) =10\frac{41}{81}\approx 10.5062$$
You could make this more exact using the error term for the trapezium rule which is $-M(b-a)/12n^2=-M/12n$ where $M$ is the average value of $f''(x)$ over the interval.
A: If you know generalized harmonic numbers
$$S_p = \sum_{n=1}^{p} \frac{1}{n^{4/5}}=H_p^{\left(\frac{4}{5}\right)}$$ Using asymptotics
$$S_p=5 p^{1/5}+\zeta \left(\frac{4}{5}\right)+\frac{1}{2}
   \frac{1}{p^{4/5}}+\cdots$$
For $p=243$, the last term is just $\frac{1}{162}$ and the first one is $15$.
Now, close to $x=1$
$$\zeta(x) \sim \frac{1}{x-1}+\gamma \implies \zeta \left(\frac{4}{5}\right) \sim \gamma-5 $$
All of the above makes
$$S_{243} \sim 10+\gamma+\frac{1}{162} $$ Since $ \gamma \sim \frac 12$ then
$$\lfloor S_{243} \rfloor =10$$
Just for your curiosity, using the above approximation, we have $S_{243} \sim 10.5834$ while the "exact" result is $10.5686$.
