# Math Help - Summation

1. ## Summation

Let be the integer closest to Find

Can anyone explain how to do this step by step? Please put in all necessary steps and all workings because I'm not too good at this.

Thanks!

2. Originally Posted by usagi_killer
Let be the integer closest to Find

Can anyone explain how to do this step by step? Please put in all necessary steps and all workings because I'm not too good at this.

Thanks!

Writing that "you are not too good at this" is not justification enough to ask for "all the work" to be done for you, since this is your homework and you must do it, so make and effort, think hard and get contented with some hints:

First, it must be clear that $f(n)>1\,\,\forall\,n\in\mathbb{N}$, so we must check first for which values of n we get $f(n)=1\Longleftrightarrow \sqrt[4]{n}\leq \frac{3}{2}=1.5\,\Longleftrightarrow n\leq \frac{81}{16}\sim 5.06$ $\Longrightarrow\,f(n)=1\,\,for\,\,n=1,2,3,4,5\,,\, \,but\,\,f(6)=2\,\,since\,,\sqrt[4]{6}=1.56>1.5$.

Next, $f(n)=2\Longleftrightarrow\,\sqrt[4]{n}\leq \frac{5}{2}=2.5\,\Longleftrightarrow n\leq \frac{625}{16}\sim 39.06$ $\Longrightarrow f(39)=2\,\,but\,\,f(40)=3\,\,since\,\,\sqrt[4]{40}\sim 2.51$...and etc.

So your sum begins $\sum\limits_{n=1}^1995\frac{1}{f(n)}=5\cdot \frac{1}{1}+24\cdot \frac{1}{2}+...$, since f(n) = 1 for 5 values (n=1,2,3,4,5), f(n) = 2 for 24 values (n= 6,7,...,39), etc.
Now you do the rest following the above method.

Tonio