Prove that $\frac{1}{1\times2013}+\frac{1}{2\times2012}+\frac{1}{3\times2011}+...+\frac{1}{2012\times2}+\frac{1}{2013\times1}<1$
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Tags: Inequality, Integers
cobbler
31.03.2015 00:03
Write $1=\frac{1}{2013}+\frac{1}{2013}+\cdots +\frac{1}{2013}$ where there are $2013$ terms. So since there are $2013$ terms on $\text{LHS}$, it follows that we can pair each term on $\text{LHS}$ with a $\frac{1}{2013}$ on $\text{RHS}$. However, the first and last terms on $\text{LHS}$ cancel with the corresponding $\frac{1}{2013}$'s on $\text{RHS}$, so it now suffices to prove that \[\frac{1}{2\times 2012}+\frac{1}{3\times 2011}+\cdots +\frac{1}{2011\times 3}+\frac{1}{2012\times 2}<\frac{1}{2013}+\frac{1}{2013}+\cdots +\frac{1}{2013}\]where there are $2013-2=2011$ terms on each side. Now, each term on $\text{LHS}$ is of the form $\frac{1}{(1+k)(2013-k)}$ where $0<k<2012$, so it suffices to show that \[(1+k)(2013-k)>2013\]\[\Leftrightarrow 2013(1+k)-k(1+k)>2013\]\[\Leftrightarrow 2013k-k(1+k)>0\]\[\Leftrightarrow 2013>1+k\]\[\Leftrightarrow 2012>k,\]which is true.
Nitzuga
31.03.2015 00:10
Note that if $m>n+1$, then $(m-1)(n+1) > mn$, so $\frac{1}{(m-1)(n+1)}<\frac{1}{mn}$ for $(m,n)=(2013,1), (2012,2), ... (1008, 1006)$. Therefor, every term in the series is less than $\frac{1}{1\times2013}$ except for itself. This means that the entire sum $$\frac{1}{1\times 2013}+\frac{1}{2\times 2012} +\frac{1}{3\times 2011}+\cdots+\frac{1}{2013\times 1}<\frac{1}{2013}+\frac{1}{2013}+\cdots +\frac{1}{2013}=1$$
(Similar to arkanm's solution.)
primesarespecial
21.12.2020 14:08
My solution is extremely different Write the nth term as 1/n*(2014-n).Taking sigma, we can use partial fractions and after splitting and reindexing the other sigma it is(2/ 2014)*H(2013);where H(n) denotes the nth harmonic number .Now it suffices to show that the H(2013)<1007.We can well approximate the Nth Harmonic Number by the natural log.Hence it suffices to show that ln(2013)+a<1007,where a is approximately the Euler-Mascheroni constant as 2013 is fairly large .Since we can Subtract 3 from both sides to "remove" the constant and need to show that ln(2013)<1003.Raise both sides to the power of e ,to get 2013<e^1003 But since e>2,e^1003>2^(1003)>2^11=2048>2013. QED.
winterrain01
21.12.2020 18:03
Notice that the largest term is $\frac{1}{1\times2013}$.
Since you could see that there are $2013$ terms, if you assume that all of the terms are $\frac{1}{1\times2013}=\frac{1}{2013}$, that means that the total sum would be $\frac{1}{2013} \times 2013 = 1$. However, since there are a lot of terms less than $\frac{1}{2013}$, that means that sum is less than 1.
agirlhasnoname
21.12.2020 18:19
winterrain01 wrote:
Notice that the largest term is $\frac{1}{1\times2013}$.
Since you could see that there are $2013$ terms, if you assume that all of the terms are $\frac{1}{1\times2013}=\frac{1}{2013}$, that means that the total sum would be $\frac{1}{2013} \times 2013 = 1$. However, since there are a lot of terms less than $\frac{1}{2013}$, that means that sum is less than 1.
Yes, I have a similar solution i(2014-i) > 2013 for all i={2,3,...,2012} so, 1/(i(2014-i)) < 1/2013 for all i={2,3,...,2012} and equality holds for i=1 & 2013 so, overall the summation on the left hand side is strictly lesser than 2013/2013 hence, LHS<1
Hamroldt
21.12.2020 19:01
We know that $1007 \cdot 1007$ is the maximum and $2013 \cdot 1$ the minimum, so $\frac{1}{2013 \cdot 1}$ is the maximum. Then the whole thing $< 2013 \cdot \frac{1}{2013} = 1$, as desired $\square$
yofro
22.12.2020 06:36
We want
$$\sum_{n=1}^{2013}\frac{1}{n(2014-n)}$$
But $\frac{1}{n(2014-n)}=\frac{1}{2014n}+\frac{1}{2014(2014-n)}$.
So this is equivalent to showing $2\sum\frac{1}{2014n}<1$ since $\sum \frac{1}{2014(2014-n)}=\sum\frac{1}{2014n}$ by symettry.
So we have:
$$\frac{2}{2014}\sum_{n=1}^{2013}\frac{1}{n}<1$$$$\sum_{n=1}^{2013}\frac{1}{n}<1007$$
Now here comes the tricky part. Notice that $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...<1(1)+2\left(\frac12\right)+4\left(\frac14\right)+...$ because $1\le 1$ and $\frac12,\frac13\le \frac12$, etc..This was motivated by the infamous proof that the harmonic series diverges.
Notice that we get all the way up to $\frac{1}{2013}<\frac{1}{1024}$.
So this is equivalent to proving the following:
$$\left(\sum_{n=1}^{11}\frac{2^n}{2^n}\right)-\sum_{n=2014}^{2048}\frac{1}{n}<1007$$$$\Leftrightarrow\left(\sum_{n=1}^{11}n\right)-\sum_{n=2014}^{2048}\frac{1}{n}<1007$$
Thus it suffices to show that
$$\sum_{n=1}^{11}n=66<1007,$$which is obviously true. $\blacksquare$
In my solution, it came down to $66<1007$ which is very weak, so I was skeptical at first. But the actual problem is very weak, see here.
Sadigly
18.01.2025 16:16
$\frac{1}{1\times2013}+\frac{1}{2\times2012}+\frac{1}{3\times2011}+...+\frac{1}{2012\times2}+\frac{1}{2013\times1}<2023\times\frac{1}{2023}=1$