Birthday special!!!

When $\left(k - \frac {1}{2}\right)^4 < n < \left(k + \frac {1}{2}\right)^4$ , then f(n) = k. Thus there are $\left \lfloor \left(k + \frac {1}{2}\right)^4 - \left(k - \frac {1}{2}\right)^4 \right\rfloor$ values of n for which f(n) = k. Expanding using the binomial theorem, $\left(k + \frac {1}{2}\right)^4 - \left(k - \frac {1}{2}\right)^4$ = $4k^3+k$

Thus, $\frac{1}{k}$ appears in the summation $4k^3 + k$ times, and the sum for each k is then $(4k^3 + k) \cdot \frac{1}{k} = 4k^2 + 1$ . From $k = 1 \to\ k = 6,$ we get $\sum_{k=1}^{6} 4k^2 + 1 = 364 + 6 = 370$ (either adding or using the sum of consecutive squares formula). But this only accounts for $\sum_{k = 1}^{6} (4k^3 + k) = 4\left(\frac{6(6+1)}{2}\right)^2 + \frac{6(6+1)}{2} = 1764 + 21 = 1785$ terms, so we still have 1995 - 1785 = 210 terms with f(n) = 7. This adds $210 \cdot \frac {1}{7} = 30$ to our summation, giving $\boxed{400}.$

I can only come up with a numerical method, using a spreadsheet.

We note that the largest $f(n)$ in this case $f(1995) = 7$ , since $\sqrt [4] {1995} = 6.6832$ .

Let $f(n) = i$ , then the largest $n$ for an $i$ is given by: $n_i = \lfloor (i + 0.5)^4 \rfloor$ , where $\lfloor x \rfloor$ is the greatest integer function. Then the number of $n$ , which $f(n) = i$ is $n_i - n_{i-1}$ , where $n_0 = 0$ and $n_7 = 1995$ .

Then, we have:

$\sum _{k=1} ^{1995} {\frac {1}{f(k)}} = \sum _{i=1} ^7 {\frac {n_i-n_{i-1}}{i}}$

The following table shows the computation of $\displaystyle \sum _{k=1} ^{1995} {\frac {1}{f(k)}}$ .

$\begin{array} {rrrcr} i & n_i & n_i-n_{i-1} & \times \frac {1}{i} = & \frac {n_i-n_{i-1}}{i} \\ \hline \\ 1 & 5 & 5 & \times \frac{1}{1} = & 5 \\ 2 & 39 & 34 & \times \frac{1}{2} = & 17 \\ 3 & 150 & 111 & \times \frac{1}{3} = & 37 \\ 4 & 410 & 260 & \times \frac{1}{4} = & 65 \\ 5 & 915 & 505 & \times \frac{1}{5} = & 101 \\ 6 & 1785 & 870 & \times \frac{1}{6} = & 145 \\ 7 & 1995 & 210 & \times \frac{1}{7} = & 30 \\ & & & \sum _{k=1} ^{1995} {\frac {1}{f(k)}} = & \boxed {400} \end{array}$

Here is my $JAVA$ code for the problem:-

Python:

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from fractions import Fraction as frac
def f(n):
    return int(round(n**(1/4.0)))
total = 0
for k in xrange(1,1996):
    total += frac(1, f(k))
print "Answer:", float(total)

And I'm a little late, but happy birthday!

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#include<iostream.h>
#include<conio.h>
#include<math.h>
int main()
{
clrscr();
double a,b,c,s;
int i;
s=0;
for (i=1;i<1996;++i)
{
b=pow(i,0.25);
c=floor(b)+0.5;
if (b<c)
{
b=floor(b);
}
else
{
b=ceil(b);
}
a=1/b;
s=s+a;
}
cout<<s;
getch();
}

Here's C++ program

define a function $g(x)$ and let $ \int g(x) dx = 6$

Here is an integer solution without floor/ceil-functions! First, collect all integers $n$ such that $f(n)=k$ . $\begin{aligned} f(n)&=k&\Leftrightarrow&&k-\frac{1}{2}<\sqrt[4]{n}&<k+\frac{1}{2}&\Leftrightarrow&&\left(k-\frac{1}{2}\right)^4<n&<\left(k+\frac{1}{2}\right)^4 \end{aligned}$ Now we need integer estimates for the left and right borders. Taking just the floor between upper and lower bound generally does not determine the number of integers in that interval (see remark below). Let $\begin{aligned} a_k&:=\left(k-\frac{1}{2}\right)^4+\frac{15}{16}=k^4-2k^3+\frac{k(3k-1)}{2}+1\in\mathbb{N}\quad\text{for any }k\in\mathbb{N} \end{aligned}$ We only need the expanded form of $a_k$ to show that it really is an integer. With the factorized form, we notice that for $n\in\mathbb{N}$ : $\begin{aligned} \left(k-\frac{1}{2}\right)^4<n&<\left(k+\frac{1}{2}\right)^4&\Leftrightarrow&&a_k\leq n&< a_{k+1}&&\text{for any }k\in\mathbb{N} \end{aligned}$

We know that $f(n)=k=const$ if and only if $a_k\leq n < a_{k+1}$ and split the sum accordingly: $\begin{aligned} \sum_{n=1}^{1995}\frac{1}{f(n)}&=\sum_{k=1}^6\frac{a_{k+1}-a_k}{k}+\sum_{n=a_7}^{1995}\frac{1}{7}=\sum_{k=1}^6\frac{\cancel{k}(4k^2+1)}{\cancel{k}}+\sum_{n=a_7}^{1995}\frac{1}{7}\\[2em] &\underset{(*)}{=}4\cdot\frac{6\cdot 7\cdot 13}{6}+6+\frac{1995-1786+1}{7}=\boxed{400} \end{aligned}$ Rem.: In the last step $(*)$ , we use the formula for the sum of squares.

Rem.: The floor function between upper and lower border of an interval generally does not return the number of integers in that interval. Let $n\in{\mathbb{Z}}$ and look at the following examples $\begin{aligned} \frac{1}{3}<n&<\frac{5}{3}&\Rightarrow &&n&=1,&\left\lfloor\frac{5}{3}-\frac{1}{3}\right\rfloor=1\\ \frac{2}{3}<n&<\frac{4}{3}&\Rightarrow &&n&=1,&\left\lfloor\frac{4}{3}-\frac{2}{3}\right\rfloor=\red{0} \end{aligned}$ In the second example, the floor function would miss the integer solution! In general, the floor function between upper and lower bound can either return the number of integers in that interval, or it is off by one, making it rather useless here!

f (n) in this question equals to {1, 2, 3, 4, 5, 6 and 7.} only.

Sum of certain number of {1, 1/ 2, 1/ 3, 1/ 4, 1/ 5, 1/ 6 and 1/ 7.} = 400

Using Excel, this question is easy. Round (Power (x, 0.25), 0) is for nearest values.