HW: Show that the series __, an n=0 converges whenever ſal < 1, and diverges whenever...
(5) Use induction to show that Ig(n) <n for all n > 1.
Soru 2. Suppose the series Ebox" converges for (x1<4. Select all that applies n=0 Yanıtınız: (-1)"b,4" converges n=0 6" M8 M 0,4" converges. n=0 61 6,6"4" diverges. n=0 00 (-1)"b,4" diverges. n=0 boxn+1 n=0 n +1 È converges for all |x<4 nb bmxn-1 converges for |x|<4 n=1 Testi duraklat Geri Sonraki
PLEASE HELP WITH PROOF!! 8. Let an > 0 for all n in 1. Show that if an converges, then Ĉ vanın converges. [Hint: Expand [van - (1/n)]2.) N =
- Let fm (x)= 7* (0<x< 1). Show that { {m} -, converges pointwise on [0, 1]. If f(x)= lim fn(x) (0<x< 1), is there an N EI such that In(x)-f(x)}< (n>N) for all x € [0, 1] simultaneously?
4. Show that for any p > 1. This shows that a p-series converges for any p > 1 р— 1 1 to an improper integral Hint: Compare Notice that the sum you need to compare n=2 2, not n 1. Find a function f(x) such that on the interval n - 1,n]. < f(x) starts at n = You might want to draw a picture to help visualize it. Think about right-hand Riemann sums with Ax= 1.
(c) A sequence {2n} satisfying 0 < In < 1/n where E(-1)"In diverges.
a < 1. Show the series on -a, a] to onverges uniformly 25.9 (a) Let 0 (b) Does the series Explain converge uniformly on (-1,1) to =0
Suppose an > 0 for n = 1,2,3,... Let An = %=1 a; for n = 1,2,3,..., Suppose &j=1 a; diverges. Show that: no aj diverges. {j=11taj wat AN an+j > 1-1 i=1 AN+) AN+k a Show that &j=1 N for k = 1,2,3,..., Hence show that I diverges. Show th: .-1 for n = 1,2,3,..., Hence show that Lj=, converges. C. An
Problem 2 Show that if the sequence of numbers (an)n-1 satisfies Inlan) < oo, then the series In ancos(nx) converges uniformly on [0, 27). This means, the partial sums Sn(x) = ) ancos(nx) define a sequence of functions {sn} = that converges uniformly on [0, 271]. Hint: First show that the sequence is Cauchy with respect to || · ||00.
bn converges 18. Let (an)n=1 and (bn)n=1 be sequences in R. Show that if and lan – an+1 < oo, then anbr converges.