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anemone
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Evaluate $\displaystyle\lower0.5ex{\mathop{\large \sum}_{n=2009}^{\infty}} \dfrac{1}{n \choose 2009}$.
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anemone said:Evaluate $\displaystyle \sum_{n=2009}^{\infty} \dfrac{1}{n \choose 2009}$.
Pranav said:Rewrite the sum as:
$$\sum_{n=2009}^{\infty} \frac{1}{n\choose 2009}=(2009)!\sum_{n=2009}^{\infty} \frac{(n-2009)!}{n!}=(2009)!\sum_{n=2009}^{\infty} \frac{1}{n(n-1)(n-2)(n-3)\cdots (n-2008)}$$
$$=(2009)!\sum_{n=0}^{\infty} \frac{1}{(n+1)(n+2)(n+3)\cdots (n+2009)}$$
To evaluate the above sum, we have to integrate the following 2009 times:
$$\frac{1}{1-x}=1+x+x^2+x^3+\cdots$$
I am unsure if the following is going to be correct. To integrate the above 2009 times, I used the technique of repeated integral as shown here: Repeated Integral -- from Wolfram MathWorld
So from above, I have to find the following:
$$\int_0^x \frac{(x-t)^{2008}}{(1-t)(2008)!}\,dt$$
at $x=1$, i.e
$$\int_0^1 \frac{(1-t)^{2007}}{(2008)!}=\frac{1}{(2008)!(2008)}$$
Hence,
$$(2009)!\sum_{n=0}^{\infty} \frac{1}{(n+1)(n+2)(n+3)\cdots (n+2009)}=(2009)!\frac{1}{(2008)!(2008)}=\boxed{ \dfrac {2009}{2008}}$$
I am not sure if what I have done is correct.
That looks like a nice approach. Thanks for sharing! :)anemone said:Observe that
$\begin{align*}\dfrac{k+1}{k} \left(\dfrac{1}{n-1 \choose k}-\dfrac{1}{n \choose k} \right)&=\dfrac{k+1}{k} \dfrac{{n \choose k}-{n-1 \choose k}}{{n \choose k}{n-1 \choose k}}\\&=\dfrac{k+1}{k} \dfrac{n-1 \choose k-1}{{n \choose k}{n-1 \choose k}}\\&=\dfrac{k+1}{k} \dfrac{(n-1)!k!k!(n-k-1)!(n-k)!}{n!(n-1)!(k-1)!(n-k)!}\\&=\dfrac{k+1}{k} \dfrac{k\cdot k! (n-k-1)!}{n!}\\&=\dfrac{1}{n-1 \choose k}\end{align*}$
An infinite sum of binomial coefficients is a mathematical expression that involves adding an infinite number of terms, where each term is a binomial coefficient. A binomial coefficient is a number that represents the number of ways to choose a subset of k elements from a set of n elements.
The most common method to evaluate an infinite sum of binomial coefficients is by using the Binomial Theorem. This theorem states that for any real or complex numbers a and b, and any non-negative integer n, the following equation holds: (a + b)^n = Σ(n, k=0) (n choose k) * a^(n-k) * b^k. This formula allows us to expand the binomial coefficients and simplify the sum.
Evaluating an infinite sum of binomial coefficients has various applications in mathematics, statistics, physics, and engineering. It is commonly used in probability theory, combinatorics, and calculus. It can also be used to solve problems related to probability distributions and series expansions.
Some common strategies for evaluating an infinite sum of binomial coefficients include using the Binomial Theorem, using properties of binomial coefficients such as Pascal's Identity, using generating functions, and using recurrence relations. These strategies can help simplify the sum and make it easier to evaluate.
Yes, there are some limitations to evaluating an infinite sum of binomial coefficients. In some cases, the sum may not have a closed form solution and may need to be approximated using numerical methods. Additionally, the complexity of the sum may increase significantly as the number of terms increases, making it difficult to evaluate.