sequences and series - Does $\sum_ {n=1}^ {\infty} (n^3 +1 )^ {1/3}-n ...
Oct 9, 2015 · Suppose I am given a infinite series as $$\sum_ {n=1}^ {\infty} (n^3 +1 )^ {1/3}-n$$ how can I tell that if it converges or diverges (by which test) , I applied D'alembert ratio test as $$\lim_ {n …
Solving a series $n(1 + n + n^2 + n^3 + n^4 +.......n^{n-1})$
This more recent post (applied to r = a = n r = a = n) answers your question: Deriving Sum of a Geometric Progression
Use mathematical induction to prove that $n^ 3 − n$ is divisible by 3 ...
Use mathematical induction to prove that n3 − n n 3 n is divisible by 3 whenever n is a positive integer. Ask Question Asked 9 years, 7 months ago Modified 7 years, 7 months ago
Proving $1^3+ 2^3 + \\cdots + n^3 = \\left(\\frac{n(n+1)}{2}\\right)^2 ...
Dec 9, 2014 · The result now follows immediately by F(n) = (n(n + 1)/2)2 ⇒ F(n) − F(n − 1) = n3 F (n) = (n (n + 1) / 2) 2 ⇒ F (n) F (n 1) = n 3 The theorem reduces the proof to a trivial mechanical …
show ${n\\brace n-2} = \\binom n3 + 3\\binom n4$ combinatorially
Dec 20, 2025 · Use combinatorial reasoning to show $$ {n\brace n-2} = \binom n3 + 3\binom n4$$ where the Stirling number is the number of partitions of $ [n]$ into $n-2$ parts.
summation - Prove that $1^3 + 2^3 + ... + n^3 = (1+ 2 + ... + n)^2 ...
This is what I've been able to do: Base case: n = 1 n = 1 L. H. S: 13 = 1 L H S: 1 3 = 1 R. H. S: (1)2 = 1 R H S: (1) 2 = 1 Therefore it's true for n = 1 n = 1. I.H ...
asymptotics - How to find $\frac {n^3} {1000} - 100n^2 - 100n + 3$ in ...
Question: Express the function n3 1000 − 100n2 − 100n + 3 n 3 1000 100 n 2 100 n + 3 in terms of the Θ notation and prove that your expression in fact fits into the Θ definition.
Mathematical induction: $9$ divides $n^3 + (n+1)^3 + (n+2)^3$
Apr 18, 2015 · Closed 10 years ago. Prove that 9 9 divides n3 + (n + 1)3 + (n + 2)3 n 3 + (n + 1) 3 + (n + 2) 3 where n n is a nonnegative integer. I have seen many questions on this site that contain the …
Series convergence test, $\sum_ {n=1}^ {\infty} \frac { (x-2)^n} {n3^n}$
Jun 28, 2020 · By the ratio test, every x value between -1 and 5 would make the series converge. we just need to find out whether x=-1, 5 makes it converge. x=-1: The series will look like this. $$\sum_ …
divisibility - Use induction to prove that $6$ divides $n^3 - n ...
Aug 1, 2016 · Now I can see that this must be true, since n3 − n = (n + 1)n(n − 1) n 3 n = (n + 1) n (n 1), i.e. the product of three consecutive integers. Therefore at least one of them will be even, and one …