Mathematics

Questions tagged [alternating-expression]

For questions related to alternating expression (or series). It is a sequence, whose terms change sign (i.e. if a term $a_n$ is positive then $a_{n+1}$ is negative and vice versa).

Basically, alternating sequence is a sequence, whose term change sign (i.e. if a term $a_n$ is positive then $a_{n+1}$ is negative and vice versa).

It is an infinite series of the form $\displaystyle \sum _{n=0}^{\infty }(-1)^{n}a_{n}$ or $\displaystyle \sum _{n=0}^{\infty }(-1)^{n+1}a_{n}$ with $a_n > 0$ for all $n$.

The signs of the general terms alternate between positive and negative. Like any series, an alternating series converges if and only if the associated sequence of partial sums converges.

For more, check out this link.

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Alternating sum of reciprocals of binomial coefficients

I'm looking for a simple proof of the identity $$ \sum_{k=0}^n \frac{(-1)^k}{\binom{n}{k}} = \frac{n+1}{n+2} (1+(-1)^n) $$ relying only on elementary properties of binomial coefficients. I obtained this by starting with the integral representation…
Dave
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Alternating sum of square root reciprocals

The alternating sum $\sum_{n=1}^{\infty}{\frac{(-1)^{n+1}}{\sqrt{n}}}=(1-\sqrt{2})\zeta(\frac{1}{2})$ according to Wolfram Alpha. I tried to derive this result myself, and found it easy using the naïve way of expanding $\zeta(\frac{1}{2})$ into its…
girobuz
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An identity involving Catalan numbers and binomial coefficients.

I stumbled upon the following identity $$\sum_{k=0}^n(-1)^kC_k\binom{k+2}{n-k}=0\qquad n\ge2$$ where $C_n$ is the $n$ th Catalan number. Any suggestions on how to prove it are welcome! This came up as a special case of a generating function for…
Cheerful Parsnip
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Combinatorial sums with binomial coefficients

Mathematica tells me that \begin{equation*} \sum_{k=a}^l \left(-\frac{1}{4}\right)^k {2k \choose k} {l \choose k} {k \choose a} = \frac{(-1)^a}{4^l} {2a \choose a} {2l - 2a \choose l-a}, \end{equation*} as well as \begin{equation*} \sum_{k=a}^l…
Josemi
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Prove $- \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sum_{k=0}^{\infty} \frac{1}{(n+k)^2} = \frac{\zeta(3)}{4}-\frac{\pi^2}{4} \log (2)$

How can I prove that $$- \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sum_{k=0}^{\infty} \frac{1}{(n+k)^2} = \frac{\zeta(3)}{4}-\frac{\pi^2}{4} \log (2)$$ The problem comes from this MSE post , where the OP wanted to solve integrals of this form $$f(x)…
user1295974
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How to prove that $2^{2n+1}=\sum_{k=0}^{2n}(-1)^{k+n}\binom{4n+2}{2k+1}\;\;\;\forall n \in\mathbb{N}$

I'm working on a related problem, which is to determine the Taylor series of $\sin(\pi x)\sinh(\pi x)$ evaluated at $x=0$ . I have succeeded in finding the series, but my coefficients are complicated sums: $\left…
Jayson Vivet
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Closed form for an alternating Euler sum with squared harmonic numbers

Let $H_n^{(p)} = 1^{-p} + 2^{-p} + \dots + n^{-p}$ be the generalized harmonic number. Consider the alternating series: $$ T = \sum_{n=1}^\infty \frac{(H_n^{(2)})^2}{n^3} (-1)^{n-1}. $$ The non-alternating version is known to be a rational…
NotEinstein
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An identity with binomial coefficients

I want to prove the following identity: if $1\leq m\leq n$ , then $$ \sum_{i=1}^{m}(-1)^{i-1}\binom{n-i}{n-m}\left[\binom{n+m}{i}-\binom{n+m}{i-1}\right]=\binom{n}{m} $$ My attempt : I wanted to use the following identity $$…
boaz
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Asymptotic Behaviour of the Remainder of Certain Alternating Series

Let $a,b >0$ be real constants. Empirical observation (as in: asking WolframAlpha) suggests $$ \lim_{n\to \infty} n \cdot \sum_{k=0}^\infty (\frac{1}{n+ak} - \frac{1}{n+b+ak}) = \frac{b}{a} \tag{$*$}$$ Note that the series in question can be…
Torsten Schoeneberg
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If $(a_n)$ is a decreasing real sequence and $\sum a_n$ converges, then does $\sum (-1)^n n a_n\ $ converge?

"Motivation"/Introduction: If $(a_n)$ is a decreasing real sequence and $\displaystyle\sum a_n $ converges, then $n a_n \to 0,\ $ for example, by the Cauchy Condensation test . If $(a_n)$ is a real sequence and $\displaystyle\sum a_n $ converges but…
Adam Rubinson
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Closed form of $\sum _{k\ge 1} \frac{(-1)^{\binom{k}{p}}}{k}$, an alternating harmonic series with the signs determined by a binomial coeffcient

In a comment to Evaluating $\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$ for $n \in \mathbb{N}$ I proposed to study this alternating harmonic sum $$s(p) = \sum _{k\ge 1}…
Dr. Wolfgang Hintze
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Alternating sum over triple products of binomial coefficients divided by $2k+1$

Let $T(n,m)$ be a rational coefficients such that $$ T(n,m) = \sum\limits_{k=0}^{m} \binom{n+m+1}{m+k+1} \binom{n+m-k}{n} \binom{m+k+1}{k} \frac{(-1)^k}{2k+1}. $$ I conjecture that for $n \geqslant 0, m \geqslant 0$ we have $$ T(n,m) =…
user929945
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Are alternating series just derivative series?

I have a thought I cannot get out of my head but also cannot seem to prove or disprove satisfyingly. Please help me clarify my thoughts. Consider an alternating series $\sum{({-1})^{n+1}u_n}$ where $u_n$ is positive. We could group two by two this…
Arno
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Gut intuition for an alternating series divergence

Here is a specific diverging alternating series I came across: $$\sum_{n=2}^{\infty} \frac{(-1)^n}{\sqrt{n}+(-1)^n}$$ This series diverges, one can show that:…
Arno
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An alternating binomial sum equal to 1

I was trying to prove that $$\sum_{j=0}^{r}(-1)^{r-j}\binom{d-j}{r-j}\sum_{l=1}^{d-k}(-1)^{l+1}\binom{d-k}{l}\binom{d-l+1}{j}=1$$ Where $0 \leq r\leq k-1$ , $k \leq \lfloor \frac{d}{2} \rfloor$ . I started by reordering the terms like that:…
Zeta16
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