Proof of sum results

I was going through some of my notes when I found both these sums with their results x0+x1+x2+x3+…=11−x,|x|<1 0+1+2x+3×2+4×3+…=1(1−x)2 I tried but I was unable to prove or confirm that these results are actually correct, could anyone please help me confirm whether these work or not? Answer 1−xn+11−x=1+x+x2+⋯+xn, now if n→∞ and |x|<1 we get … Read more

Find sum with binomial coefficients and powers of 2

Find this sum for positive n and m: S(n, m) = \sum_{i=0}^n \frac{1}{2^{m+i+1}}\binom{m+i}{i} + \sum_{i=0}^m \frac{1}{2^{n+i+1}}\binom{n+i}{i}. Obviosly, S(n,m)=S(m,n). Therefore I’ve tried find T(n,m) = \sum_{i=0}^n \frac{1}{2^{m+i}}\binom{m+i}{i} by T(n, m+1), but in binomial we have \binom{m+i+1}{i} = \binom{m+i}{i} + \binom{m+i}{i-1}, and this “i-1” brings nothing good. Other combinations like T(n+1,m+1)+T(n,m) also doesn’t provide advance. Any ideas? … Read more

Determine the Value of ∑∞n=0(1+n)xn\sum_{n=0}^{\infty} (1+n)x^n [duplicate]

This question already has answers here: How can I evaluate ∑∞n=0(n+1)xn? (23 answers) Closed 6 years ago. For x∈R with |x|<1. Find the value of ∞∑n=0(1+n)xn Answer ∞∑n=0(1+n)xn=∞∑n=0xn+∞∑n=0nxn 11−x+xddx(11−x)=11−x+x(1−x)2=1(1−x)2 AttributionSource : Link , Question Author : gaufler , Answer Author : E.H.E

Why does this sum equal zero?

Let $\gamma$ be a piece-wise, smooth, closed curve. Let $[t_{j+1}, t_{j}]$ be an interval on the curve. Prove, $$\int_{\gamma} z^m dz=0$$ In the proof it states $$\int_{t_{j}}^{t_{j+1}} \gamma^m(t) \gamma'(t)=\frac{1}{m+1} [\gamma^{m+1}(t_{j+1})-\gamma^{m+1}(t_{j})]$$ Which I can see why. Next it claims, $$\int_{\gamma} z^m dz=\sum_{j=0}^{n-1} [\frac{1}{m+1} [\gamma^{m+1}(t_{j+1})-\gamma^{m}(t_j)]]=\frac{1}{m+1} [\gamma^{m+1}(b)-\gamma^{m+1}(a)]=0$$ I understand why it is $0$ but how did they … Read more

Is ∑∞n=1ansin(nx)\sum_{n=1}^\infty a_n\sin(nx) converges on [ε,2π−ε][\varepsilon, 2\pi-\varepsilon]?

Let an, a sequence monotonically decreasing to 0. Consider ∞∑n=1ansin(nx) Is the series converges uniformly on [ε,2π−ε]? (ε>0) Basically we could use Dirichlet’s test. We want to show that ∑∞n=1sin(nx) is bounded. Indeed: ∞∑n=1sin(nx)=i2(∞∑n=1(eix)n+∞∑n=1(e−ix)n)≤i2(11−eix+11−e−ix)≤11−ei(2π−ε)<∞ BUT, clearly, g(π2)=∞∑n=1sinnπ2=∞ Where is the mistake? Answer We show that if an is monotonically decreasing, then the series ∞∑n=1ansin(nx) is … Read more

Prove or disprove: ∑b∨d=xτ(b)τ(d)=τ(x)3 \sum_{b \vee d = x} \tau(b) \tau(d) = \tau(x)^3

Can somebody prove or disprove? Let τ be the divisors function, so that τ(6)=#{1,2,3,6}=4 ∑b∨d=xτ(b)τ(d)=τ(x)3 Here I am using b∨d=lcm(b,d) since it is the join of two numbers in the multiplicative lattice (N,×). This statement seems to be true. Let’s try x=6. The left hand side is: 2×τ(6)[τ(1)+τ(2)+τ(3)]+τ(6)2+τ(3)[τ(2)+τ(6)]+τ(2)[τ(3)+τ(6)]=64 and indeed τ(6)3=64. One option is to … Read more

How do you find the condition where the Cauchy-Schwarz inequality is equal?

Cauchy-Schwarz inequality: |n∑i=1aibi|2≤n∑i=1|ai|2n∑i=1|bi|2 The answer is known to be when aik+bi=0 for some k∈R, or any other equivalence (e.g. in linear algebra, when the vectors are linearly dependent). My question is, without a priori knowledge that equality holds for Cauchy-Schwarz if and only if [condition such as aik+bi=0 or some equivalent condition], how do you … Read more

Changing signs to minus signs to obtain a sum of zero.

Consider the sum 1+2+3+…+101. Is it possible to change some of the plus signs to minus signs so that the sum is zero? Well, I know by using Gauss’ method 1+2+…+100=5050 then 5050+101=5151 So I started to see if I can find a pattern but Im not sure. Here is what I did: I aligned … Read more

Proving that $\sum_{i=1}^n\frac{1}{i^2}<2-\frac1n$ for $n>1$ by induction [duplicate]

This question already has answers here: Proving $1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$ for all $n\geq 2$ by induction (5 answers) Closed 3 years ago. Prove by induction that $1 + \frac {1}{4} + \frac {1}{9} + … +\frac {1}{n^2} < 2 – \frac{1}{n}$ for all $n>1$ I got up to using the inductive hypothesis to prove that … Read more