Find the integer part of $$\sum^{9999}_{k=1}\frac{1}{\sqrt[4]{k}}.$$
Is there any way without using definite integration or telescopic series?
What I tried: integration means summation and we have $$\sum^{9999}_{k=1}\frac{1}{\sqrt[4]{k}}\approx \int^{10,000}_{1}\frac{1}{x^{\frac{1}{4}}}dx=\frac{4}{3}\cdot 1000$$ How do I solve it? Help me for that problem. Thanks.
Yes, integration is the right tool to find sufficiently precise bounds for the given summation. For $x>1$, we have that $$\frac1{\sqrt[4]{x}}\le\frac1{\sqrt[4]{\lfloor x\rfloor}}<\frac1{\sqrt[4]{x-1}}$$ and therefore $$1332=\int_1^{10000}\frac{dx}{\sqrt[4]{x}}<\sum_{n=1}^{9999}\frac1{\sqrt[4]{n}}<1+\int_2^{10000}\frac{dx}{\sqrt[4]{x-1}}=1+\int_1^{9999}\frac{dx}{\sqrt[4]{x}}<1+1332.$$ Hence the integer part is $1332$.