Chaitanya's Random Pages

June 29, 2014

Melbourne’s late start to cooler temperatures in 2014

Filed under: climate and weather — ckrao @ 11:21 am

It occurred to me that Melbourne has had an unusual number of cooler days in 2014 up to the middle of June. After some checking here, I found that from January 1 to June 17 this year, the Melbourne Regional Office has only twice recorded days with a maximum temperature less than 15°C (May 2 [14.3°C] and May 4 [14.2°C]). This beats 2003 when the previous fewest number of 5 such instances took place prior to June 18. The graph and table below show that in the past this number has been as high as 40, but in recent years there is a clear downward trend.


Melbourne Regional Office Average number of sub-15°C days before June 18
Average 1856-2013 21.6
Average 1971-2013 15.0
Average 2004-2013 11.7
2014 2


If we do the same calculation for a station further away from the city centre, say the international airport just over 20km away, there is not as much historical data and the effect is not as pronounced, but this year still equals the previous record of 2003 (11 times). In recent times the sub-15°C days prior to June 18 have been about twice as frequent here as the Regional Office.


Melbourne Airport Average number of sub-15°C days before June 18
Average 1971-2013 24.7
Average 2004-2013 23.1
2014 11

The data is from Australia’s Bureau of Meteorology website.

June 28, 2014

A few sums involving inverses of binomial coefficients

Filed under: mathematics — ckrao @ 11:11 am

One of the problems from the 1990 Australian Mathematical Olympiad ([1]) asks one to prove the following result.

\displaystyle \sum_{k=1}^{2n-1}\frac{(-1)^{k-1}}{\binom{2n}{k}} = \frac{1}{n+1}.\quad\quad (1)

One solution in [1] makes use of the following identity:

\begin{aligned}  \frac{1}{\binom{m}{k}} &= \frac{k!(m-k)!}{m!}\\  &= \frac{k!(m-k)!(m+1)(m+2)}{m!(m+1)(m+2)}\\  &= \left(\frac{m+1}{m+2}\right)\frac{k!(m-k)!(m+2)}{(m+1)!}\\  &= \left(\frac{m+1}{m+2}\right)\frac{k!(m-k)!((m-k+1) + (k+1))}{(m+1)!}\\  &=\left(\frac{m+1}{m+2}\right)\frac{k!(m-k+1)! + (k+1)!(m-k)!}{(m+1)!}\\  &= \frac{m+1}{m+2}\left(\frac{1}{\binom{m+1}{k}}+ \frac{1}{\binom{m+1}{k+1}}\right).\quad\quad(2)  \end{aligned}

Then setting m=2n, (1) becomes a telescoping sum:

\begin{aligned}  \sum_{k=1}^{2n-1}\frac{(-1)^{k-1}}{\binom{2n}{k}}  &= \sum_{k=1}^{2n-1} \frac{2n+1}{2n+2}\left(\frac{(-1)^{k-1}}{\binom{2n+1}{k}}+ \frac{(-1)^{k-1}}{\binom{2n+1}{k+1}}\right)\\  &= \frac{2n+1}{2n+2}\left( \sum_{k=1}^{2n-1} \frac{(-1)^{k-1}}{\binom{2n+1}{k}} + \sum_{k=1}^{2n-1} \frac{(-1)^{k-1}}{\binom{2n+1}{k+1}}\right)\\  &= \frac{2n+1}{2n+2}\left( \frac{(-1)^0}{\binom{2n+1}{1}} + \sum_{k=2}^{2n-1} \frac{(-1)^{k-1}}{\binom{2n+1}{k}} + \frac{(-1)^{2n-1-1}}{\binom{2n+1}{2n-1+1}} + \sum_{k=1}^{2n-2} \frac{(-1)^{k-1}}{\binom{2n+1}{k+1}} \right)\\  &= \frac{2n+1}{2n+2}\left( \frac{1}{2n+1} + \frac{1}{2n+1} + \sum_{k=1}^{2n-2} \frac{(-1)^{k}}{\binom{2n+1}{k+1}} + \sum_{k=1}^{2n-2} \frac{(-1)^{k-1}}{\binom{2n+1}{k+1}} \right)\\  &= \frac{2n+1}{2n+2} \left(\frac{2}{2n+1} + 0\right)\\  &= \frac{1}{n+1}.  \end{aligned}

Using a similar argument one can prove the more general identity

\displaystyle \sum_{k=0}^n\frac{(-1)^k}{\binom{n}{k}} = \frac{n+1}{n+2}\left(1 + (-1)^n\right).\quad\quad (3)

How about the sum \displaystyle S(n) := \sum_{k=0}^{n}\frac{1}{\binom{n}{k}}? Using (1) we find the recursion

\begin{aligned}  S(n) &= \frac{n+1}{n+2}\left(\sum_{k=0}^n \frac{1}{\binom{n+1}{k}} + \frac{1}{\binom{n+1}{k+1}}\right)\\  &= \frac{n+1}{n+2}\left(S(n+1)-1 + S(n+1)-1\right)\\  &= \frac{2(n+1)}{n+2}\left(S(n+1)-1 \right). \quad\quad (4)\\  \end{aligned}

Hence we have the recurrence relation

\displaystyle S(n+1) = \frac{n+2}{2(n+1)}S(n) + 1.\quad\quad(5)

From this S(n) does not have a closed form solution but using induction on n we can prove the following relations found in [2].

\displaystyle S(n) := \sum_{k=0}^{n}\frac{1}{\binom{n}{k}} = \frac{n+1}{2^{n+1}}\sum_{k=1}^{n+1}\frac{2^k}{k} = (n+1)\sum_{k=0}^n \frac{1}{(n+1-k)2^k}.\quad\quad(6)

Firstly note that when n=0, the three expressions in (5) become \frac{1}{\binom{0}{0}},  \frac{1}{2^{1}}\frac{2^1}{1} and \frac{1}{1}, which are all equal to 1. Hence (6) holds for n=0. Assume it holds for n =m. That is,

\displaystyle S(m) := \sum_{k=0}^{m}\frac{1}{\binom{m}{k}} = \frac{m+1}{2^{m+1}}\sum_{k=1}^{m+1}\frac{2^k}{k} = (m+1)\sum_{k=0}^m \frac{1}{(m+1-k)2^k}.\quad\quad(7)

Then using (5), on the one hand,

\begin{aligned}  S(m+1) &= 1 + \frac{m+2}{2(m+1)}S(m)\\  &= 1 + \frac{m+2}{2(m+1)} \frac{m+1}{2^{m+1}}\sum_{k=1}^{m+1}\frac{2^k}{k} \\  &= 1 + \frac{m+2}{2^{m+2}}\sum_{k=1}^{m+1}\frac{2^k}{k}\\  &= \frac{(m+2)2^{m+2}}{2^{m+2}(m+2)} + \frac{m+2}{2^{m+2}}\sum_{k=1}^{m+1}\frac{2^k}{k}\\  &= \frac{m+2}{2^{m+2}}\sum_{k=1}^{m+2}\frac{2^k}{k},\quad\quad(8)  \end{aligned}

while on the other,

\begin{aligned}S(m+1) &= 1 + \frac{m+2}{2(m+1)}S(m)\\  &= 1 + \frac{m+2}{2(m+1)} (m+1)\sum_{k=0}^m \frac{1}{(m+1-k)2^k}\\  &= 1 + (m+2)\sum_{k=0}^m \frac{1}{(m+1-k)2^{k+1}}\\  &= 1 + (m+2)\sum_{k=0}^m \frac{1}{(m+2-(k+1))2^{k+1}}\\  &= 1 + (m+2)\sum_{k=1}^{m+1} \frac{1}{(m+2-k)2^{k}}\\  &= (m+2)\sum_{k=0}^{m+1} \frac{1}{(m+2-k)2^k}.\quad \quad(9)  \end{aligned}

Equations (8) and (9) show that if (6) holds for n=m, then it does for n=m+1. By the principle of mathematical induction, (6) then is true for all integers n\geq 0.

Another related sum that can be expressed in terms of S(n) is \displaystyle \frac{1}{\binom{2n}{0}} + \frac{1}{\binom{2n}{2}} + \frac{1}{\binom{2n}{4}} + \ldots + \frac{1}{\binom{2n}{2n}}. We have

\begin{aligned}& \sum_{k=0}^{2n} \frac{1}{\binom{2n}{2k}}\\&= \frac{2n+1}{2n+2}\sum_{k=0}^{2n} \frac{1}{\binom{2n+1}{k}}\text{\quad \quad (by (5))}\\  &= \frac{2n+1}{2n+2} S(2n+1).\quad\quad(10)  \end{aligned}

Note that more identities with inverses of binomial coefficients are in [2] (and references therein), where the integral

\displaystyle \frac{1}{\binom{n}{k}} = (n+1) \int_0^1 t^k (1-t)^{n-k}\ \text{d}t

is utilised.


[1] A. W. Plank, Mathematical Olympiads: The 1990 Australian Scene, University of Canberra, 1990.

[2] T. Mansour, Combinatorial Identities and Inverse Binomial Coefficients, available at

Create a free website or blog at

%d bloggers like this: