Question

About infinite series converge or diverge thx

Answer 1

\[\sum_{k=0}^{\infty}[e ^{-k}-e ^{-(k+1)}]\] converge or diverge, if converge what is it converge to?

Answer 2

what class is it?

Answer 3

AP Calc BC

Answer 4

\[e^{-k} - e^{-(k+1)} = e^{-k} (1 - \frac{1}{e})\]

Answer 5

then use integral test?

Answer 6

yes that's one option.

Answer 7

what is the best option?

Answer 8

\[\left(E^0-E^{-1}\right)+\left(E^{-1}-E^{-2}\right)+\left(E^{-2}-E^{-3}\right)+\text{...}.+\left(E^{-(k-1)}-E^{-(k)}\right)\]

Answer 9

oh, i see

Answer 10

\[E^0 - E^{-k}\]

Answer 11

if you are solving for series then I will say that integral test is the best option except that for the formula the summation starts from n = 1 so I'm not sure if we can use it in this case where its from 0 to infinity.
\[\sum_{n =1}^{\infty}a_{n} (and) \int\limits_{1}^{\infty} f(x) dx\]

Answer 12

if f is positive, continuous, and decreasing for x greater than or equal to 1 and \[a_{n} = f(n)\]

Answer 13

could use limit instead

Answer 14

?

Answer 15

The series will converge to 1

Answer 16

based on imranmeah91's method

Answer 17

yes i got the same answer

Answer 18

what about this problem

Answer 19

It is 1,

\[\sum _{k=0}^{\infty } \left(E^{-k}-E^{-(k+1)}\right)\]=1



\[\sum _{k=0}^{\infty } \left(E^{-k}\right)\text{ }-E^{-(k+1)}=\frac{e}{-1+e}-e^{-1-k}\]

Answer 20

what about this problem is it converge or diverge?\[\sum_{k=0}^{\infty}(-1)^{k+1}\times(4/3^{k})\]

Answer 21

thx

Answer 22

this is alternating series, so we have to check for both absolute convergence, and conditional convergence

Answer 23

since 4/3 > 1 , it is not absolutely convergent

Answer 24

but my teacher told me it is converge to -3, i have a disagreement with her

Answer 25

http://tinyurl.com/3bxnq3h

Answer 26

wolfram disagrees with your teacher

Answer 27

oh, the k is only the power of 3

Answer 28

\[\sum _{k=0}^{\infty } (-1)^{1+k} \left(\frac{4}{3^k}\right)\]

this?

Answer 29

yes

Answer 30

i think my teacher is right then

Answer 31

yes, she is right

\[\underset{k=0}{\overset{\infty }{4\sum }}(-1)^{1+k} \left(\frac{1}{3}\right)^k\]

Answer 32

is there a way could use integral test to get the value of it?

Answer 33

no, integral test only determines if whether the series converges, not what it converges to

Answer 34

so have to use the limit of the sum?

Answer 35

but how to get the formula of the sum of these kind of serious?

Answer 36

Unless there are easy tricks (like for above series), it is difficult to find what a series converges too.

Answer 37

ok, thank you

Answer 38

but how suppose i can figure this out by hand? not by the Wolfram..

Answer 39

\[4\sum _{k=0}^{\infty } \left(-\frac{1}{3}\right)^k(-1)^{k+1}\]

can be written as
\[4\left(-\left(\frac{1}{3}\right)^0+\left(\frac{1}{3}\right)^1-\left(\frac{1}{3}\right)^2+\text{...}\text{..}+\left(\frac{1}{3}\right)^k\right)\]

multiply by that by -1

\[4\left(\left(\frac{1}{3}\right)^0-\left(\frac{1}{3}\right)^1+\left(\frac{1}{3}\right)^2+\text{...}\text{..}+\left(\frac{1}{3}\right)^k\right)\]


\[-4\sum _{k=0}^{\infty } \left(-\frac{1}{3}\right)^k\]
=
\[4\sum _{k=0}^{\infty } \left(\frac{1}{3}\right)^k(-1)^{k+1}\]


\[-4\sum _{k=0}^{\infty } \left(-\frac{1}{3}\right)^k\]= using geometric series it is -3

Answer 40

ok, i see thank you so much

Answer 41

wow, AP class got tougher than when I took it