<< Chapter < Page Chapter >> Page >

n = 1 1 n 13 1 ( n + 1 ) 13

Got questions? Get instant answers now!

n = 1 ( n n + 1 )

S k = 1 k + 1 diverges

Got questions? Get instant answers now!

n = 1 ( sin n sin ( n + 1 ) )

Got questions? Get instant answers now!

Express the following series as a telescoping sum and evaluate its n th partial sum.

n = 1 ln ( n n + 1 )

n = 1 ln n ln ( n + 1 ) , S k = ln ( k + 1 )

Got questions? Get instant answers now!

n = 1 2 n + 1 ( n 2 + n ) 2 ( Hint: Factor denominator and use partial fractions.)

Got questions? Get instant answers now!

n = 2 ln ( 1 + n 1 ) ln n ln ( n + 1 )

a n = 1 ln n 1 ln ( n + 1 ) and S k = 1 ln ( 2 ) 1 ln ( k + 1 ) 1 ln ( 2 )

Got questions? Get instant answers now!

n = 1 ( n + 2 ) n ( n + 1 ) 2 n + 1 ( Hint: Look at 1 / ( n 2 n ) . )

Got questions? Get instant answers now!

A general telescoping series is one in which all but the first few terms cancel out after summing a given number of successive terms.

Let a n = f ( n ) 2 f ( n + 1 ) + f ( n + 2 ) , in which f ( n ) 0 as n . Find n = 1 a n .

n = 1 a n = f ( 1 ) f ( 2 )

Got questions? Get instant answers now!

a n = f ( n ) f ( n + 1 ) f ( n + 2 ) + f ( n + 3 ) , in which f ( n ) 0 as n . Find n = 1 a n .

Got questions? Get instant answers now!

Suppose that a n = c 0 f ( n ) + c 1 f ( n + 1 ) + c 2 f ( n + 2 ) + c 3 f ( n + 3 ) + c 4 f ( n + 4 ) , where f ( n ) 0 as n . Find a condition on the coefficients c 0 ,… , c 4 that make this a general telescoping series.

c 0 + c 1 + c 2 + c 3 + c 4 = 0

Got questions? Get instant answers now!

Evaluate n = 1 1 n ( n + 1 ) ( n + 2 ) ( Hint: 1 n ( n + 1 ) ( n + 2 ) = 1 2 n 1 n + 1 + 1 2 ( n + 2 ) )

Got questions? Get instant answers now!

Evaluate n = 2 2 n 3 n .

2 n 3 1 = 1 n 1 2 n + 1 n + 1 , S n = ( 1 1 + 1 / 3 ) + ( 1 / 2 2 / 3 + 1 / 4 ) + ( 1 / 3 2 / 4 + 1 / 5 ) + ( 1 / 4 2 / 5 + 1 / 6 ) + = 1 / 2

Got questions? Get instant answers now!

Find a formula for n = 1 1 n ( n + N ) where N is a positive integer.

Got questions? Get instant answers now!

[T] Define a sequence t k = n = 1 k 1 ( 1 / k ) ln k . Use the graph of 1 / x to verify that t k is increasing. Plot t k for k = 1 100 and state whether it appears that the sequence converges.

t k converges to 0.57721 t k is a sum of rectangles of height 1 / k over the interval [ k , k + 1 ] which lie above the graph of 1 / x .
This is a graph of a sequence in quadrant one that begins close to 0 and appears to converge to 0.57721.

Got questions? Get instant answers now!

[T] Suppose that N equal uniform rectangular blocks are stacked one on top of the other, allowing for some overhang. Archimedes’ law of the lever implies that the stack of N blocks is stable as long as the center of mass of the top ( N 1 ) blocks lies at the edge of the bottom block. Let x denote the position of the edge of the bottom block, and think of its position as relative to the center of the next-to-bottom block. This implies that ( N 1 ) x = ( 1 2 x ) or x = 1 / ( 2 N ) . Use this expression to compute the maximum overhang (the position of the edge of the top block over the edge of the bottom block.) See the following figure.

This is a diagram of the edge of a table with several blocks stacked on the edge. Each block is pushed slightly to the right, and over the edge of the table. An arrow is drawn with arrowheads at both ends from the edge of the table to a line drawn down from the edge of the highest block.
Got questions? Get instant answers now!

Each of the following infinite series converges to the given multiple of π or 1 / π .

In each case, find the minimum value of N such that the N th partial sum of the series accurately approximates the left-hand side to the given number of decimal places, and give the desired approximate value. Up to 15 decimals place, π = 3.141592653589793... .

[T] π = −3 + n = 1 n 2 n n ! 2 ( 2 n ) ! , error < 0.0001

N = 22 , S N = 6.1415

Got questions? Get instant answers now!

[T] π 2 = k = 0 k ! ( 2 k + 1 ) ! ! = k = 0 2 k k ! 2 ( 2 k + 1 ) ! , error < 10 −4

Got questions? Get instant answers now!

[T] 9801 2 π = 4 9801 k = 0 ( 4 k ) ! ( 1103 + 26390 k ) ( k ! ) 4 396 4 k , error < 10 −12

N = 3 , S N = 1.559877597243667 ...

Got questions? Get instant answers now!

[T] 1 12 π = k = 0 ( −1 ) k ( 6 k ) ! ( 13591409 + 545140134 k ) ( 3 k ) ! ( k ! ) 3 640320 3 k + 3 / 2 , error < 10 −15

Got questions? Get instant answers now!

[T] A fair coin is one that has probability 1 / 2 of coming up heads when flipped.

  1. What is the probability that a fair coin will come up tails n times in a row?
  2. Find the probability that a coin comes up heads for the first time after an even number of coin flips.

a. The probability of any given ordered sequence of outcomes for n coin flips is 1 / 2 n . b. The probability of coming up heads for the first time on the n th flip is the probability of the sequence T T T H which is 1 / 2 n . The probability of coming up heads for the first time on an even flip is n = 1 1 / 2 2 n or 1 / 3 .

Got questions? Get instant answers now!
Practice Key Terms 7

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Calculus volume 2' conversation and receive update notifications?

Ask