- FIND THE RADIUS OF 
CONVERGENCE OF A POWER SERIES.
NOW HERE IS A POWER SERIES,
AND WE WANT TO FIND THE 
INTERVAL OF x VALUES
FOR WHICH THIS SERIES 
CONVERGES.
FOR EXAMPLE, 
FOR AN x VALUE OF 1,
IF YOU PUT THAT IN THERE 
FOR x, YOU'RE GOING TO GET
5 TO THE n POWER, AND SO THAT 
SERIES DIVERGES HORRIBLY.
SO x = 1 IS NOT GOING 
TO BE A VALUE
THAT MAKES 
THE SERIES CONVERGE.
SO TO FIND THE VALUES 
FOR WHICH IT DOES CONVERGE,
WE'RE GOING TO USE 
OUR RATIO TEST. OKAY?
SO AGAIN, THE RATIO TEST SAYS 
YOU TAKE THE ABSOLUTE VALUE
AND RATIO OF SUCCESSIVE TERMS 
OF THE SERIES
AND IF THAT LIMITS OUT TO L 
AND IF L < 1,
THEN THE SERIES CONVERGES.
L > 1 OR IF L IS INFINITE, 
THEN THE SERIES DIVERGES.
L = 1, TEST INCONCLUSIVE.
SO LET'S SEE.
LET'S TAKE OUR SERIES 
AND APPLY THE RATIO TEST.
HERE'S A SERIES RATIO TEST,
An + 1/An ABSOLUTE VALUE,
PLUG IN WHAT IT IS, 
(5x) TO THE n + 1 POWER,
DIVIDED BY ABSOLUTE VALUE (5x) 
TO THE n.
THAT GETS OVER TO 
HERE LIKE THAT,
AND THEN JUST CANCEL THOSE 
OFF, AND YOU GET 5x.
AND THIS n GOES TO INFINITY, 
ABSOLUTE VALUE OF 5x = 5x.
OKAY. WE WANT TO SEE
WHAT x VALUES MAKE THE SERIES 
CONVERGE.
OUR RATIO TEST SAYS THAT THE 
SERIES WILL CONVERGE
IF THAT RESULT IS < 1, 
SO WE SOLVE THAT OVER
AND WE GET THIS STATEMENT.
AND THAT SAYS THAT OUR RADIUS 
OF CONVERGENCE IS 1/5,
AND HERE'S OUR INTERVAL 
OF CONVERGENCE AT LEAST,
BUT IT MIGHT ALSO CONVERGE AT 
THE ENDPOINTS.
WE'VE GOT TO CHECK THEM OUT.
SO WHAT ABOUT THE BOUNDARIES?
THESE TWO HERE.
SO PLUG THOSE IN. IF x = 1/5, 
PLUG THAT IN HERE,
AND YOU'RE GOING TO GET 1 
TO THE n
AND A SIMILAR 1 TO THE n 
DIVERGES,
SO THIS IS OUT.
WHAT ABOUT -1/5?
THEN PUT IT IN HERE, 
YOU GET THIS.
THEN IT'S THIS, 
AND THAT DIVERGES
BECAUSE THE PARTIAL SUMS OF Sn 
OSCILLATE BETWEEN 1 AND 0.
SO IT'S 1, THEN 0, 
THEN 1, THEN 0.
SO THAT LIMIT DOES NOT EXIST, 
HENCE THE SERIES DIVERGES.
SO THE ENTIRE INTERVAL HAPPENS 
TO BE JUST THESE VALUES HERE,
AND THE RADIUS OF CONVERGENCE 
WAS THIS.
HERE'S ANOTHER EXAMPLE, 
THIS ONE RIGHT HERE.
SO WE DO OUR RATIO TEST.
AND YOU PUT THE n + 1 IN 
THERE, OF COURSE,
AND YOU WORK IT OUT.
GET TO HERE.
AND THEN WE SAY, OKAY,
THE LIMIT OF THAT, 
WHAT DOES IT LIMIT TO?
SO THIS IS HERE.
WE FACTOR OUT 4x - 8 
OUT OF THIS ALIGNMENT,
BECAUSE n IS MOVING, 
NOT THE x. THE x IS FIXED
FOR A PARTICULAR x VALUE.
IT COULD TAKE ON ANY VALUE, 
BUT ONCE IT'S TAKEN ON,
IT'S FIXED.
IT'S THE n THAT'S MOVING.
O THAT EQUALS THIS.
AND AS n GOES TO INFINITY, 
THIS HERE GOES TO (2),
AND WE WANT THIS LIMIT TO BE 
< 1 FOR OUR SERIES TO CONVERT.
SO THESE STEPS HERE 
ARE SOLVING IT OUT.
SO THIS SHOWS THAT THE RADIUS 
OF CONVERGENCE ABOUT X = 2
IS R = 1/8 CENTERED AT X = 2.
IT MEANS THAT THE SERIES 
CONVERGES FOR AT LEAST
THESE VALUES HERE, WHICH MEANS 
THIS, WHICH MEANS THESE.
BUT WHAT ABOUT THE ENDPOINTS?
FOR x = 15/8, 
YOU'RE SERIES TERM IS THIS.
PUT IN THE 15/8 FOR x.
WORK IT OUT, WORK IT OUT, 
WORK IT OUT TO HERE, TO THERE.
AND OUR SERIES HERE 
THEN EQUALS THIS SERIES,
BECAUSE ALL THIS STUFF HERE 
WAS EQUAL TO THIS.
AND THIS SERIES HERE IS THE 
ALTERNATING HARMONIC SERIES,
WHICH CONVERGES.
SO THE SERIES CONVERGES 
FOR x = 15/8.
WHAT ABOUT x = 17/8?
THEN PUT IN (17/8).
WORK IT ALL OUT.
IT COMES OUT TO 1/n.
SO THE SERIES WITH THE (17/8)
IN THERE EQUALS THIS 
SERIES HERE.
THAT'S THE REGULAR 
HARMONIC SERIES.
IT DIVERGES.
HENCE THE INTERVAL 
OF CONVERGENCE
FOR THE ORIGINAL SERIES IS 
15/8 < OR EQUAL TO x < 17/8.
YOU GET THE 15/8.
YOU DON'T GET THE 17/8.
YOU GET ANY NUMBER 
IN BETWEEN THERE.
