# Approximating the Sum of a Convergent Series

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1 Approximatig the Sum of a Coverget Series Larry Riddle Ages Scott College Decatur, GA The BC Calculus Course Descriptio metios how techology ca be used to explore covergece ad divergece of series, ad lists various tests for covergece ad divergece as topics to be covered. But o specific metio is made of actually estimatig the sum of a series, ad the oly discussio of error bouds is for alteratig series ad the Lagrage error boud for Taylor polyomials. With just a little additioal effort, however, studets ca easily approximate the sum of may commo coverget series ad determie how precise that approximatio will be. Approximatig the Sum of a Positive Series Here are two methods for estimatig the sum of a positive series whose covergece has bee established by the itegral test or the ratio test. Some fairly weak additioal requiremets are made o the terms of the series. Proofs are give i the appedix. Let S ad let the th partial sum be S a k.. Suppose f() where the graph of f is positive, decreasig, ad cocave up, ad the improper itegral f(x) dx coverges. The S + + f(x) dx + + k < S < S + f(x) dx +. () (If the coditios for f oly hold for x N, the iequality () would be valid for N.) +. Suppose ( ) is a positive decreasig sequece ad lim L <. If + If + decreases to the limit L, the ( ) L S + L icreases to the limit L, the S < S < S () < S < S + ( ) L. (3) L

2 Example : S The fuctio f(x) is positive with a graph that is decreasig ad cocave up for x, ad x f() for all. I additio, f(x) dx coverges. This series coverges by the itegral test. By iequality (), S ( + ) < S < S + ( + ). (4) This iequality implies that S is cotaied i a iterval of width ( + ) + ( + ). If we wated to estimate S with error less tha 0.000, we could use a value of with < (+) ad the take the average of the two edpoits i iequality (4) as a approximatio for S. The table feature o a graphig calculator shows that 7 is the first value of that works. Iequality (4) the implies that < S < ad a reasoable approximatio would be S.645 to three decimal places. With 00, iequality (4) actually shows that < S < , ad hece we kow for sure that S Of course, i this case we actually kow that S π Notice also that S , so the partial sum with 00 terms is a poor approximatio by itself. Example : S Let f(x) 4 + x. The graph of f is decreasig ad cocave up for x. Also x 4 + x x 4 + π 4 arcta( ) ad so the improper itegral coverges. We ca therefore use iequality () for, ad so S + π 4 arcta(( + ) ) + + (( + ) 4 + ) < S < S + π 4 + arcta( ) (( + ) 4 + ). for. Usig 0 i this iequality yields < S < We ca coclude that S to three decimal places. Example 3: S 0! The terms of this series are decreasig. I additio, + ( + )!! + We will use the covetio for positive edpoits of trucatig the left edpoit of the iterval ad roudig up the right edpoit. This will make the iterval slightly larger tha that give by the actual symbolic iequality.

3 which decreases to the limit L 0. By iequality () S < S < S + (+)! + S +!. for all. Usig 0 i this iequality yields.7888 < S <.7889 ad hece S These, of course, are the first seve decimal places of e Example 4: S We have 5 + ( + ) which icreases to the limit L 5. Accordig to iequality (3) which simplifies to S + S + (+) ( + ( ) + 5 ) < S < S + 5 ( ) 5 < S < S With 5, this iequality shows that < S < Example 5: S We have! + ( ) ( + )! ( + ) +! ( + which is less tha for all ad which decreases to the limit L e. From iequality () we get (after some simplificatio) Usig 0 gives < S < S +! e < S < S! + ( + ). Approximatig the Sum of a Alteratig Series Let S ( ) + ad let the th partial sum be S ( ) k+ a k. We assume that ( ) is a positive decreasig sequece that coverges to 0.. The stadard error boud is give by k S + < S < S + + (5) ) 3

4 . Suppose the sequece defied by b + decreases mootoically to 0. (Oe way to achieve this is if f() where f is positive with a graph that is decreasig asymptotically to 0 ad cocave up.) The if S < S, the S + + < S < S + ; (6) if S < S, the S < S < S +. (7) Both of these ca be summarized by the iequality + < S S <. Iequality (5) is credited to Leibiz ad is the error boud described i the BC Calculus Course Descriptio. Iequalities (6) ad (7) are cosequeces of a proof published i 96 by Philip Calabrese, the a udergraduate studet at the Illiois Istitute of Techology (see referece []). Calabrese proved that S S < ɛ if ɛ, ad that furthermore, if ɛ for some, the S is the first partial sum withi ɛ of the sum S. See the appedix for the derivatio of iequalities (6) ad (7). Example 6: S ( ) + 4 This is a alteratig series that coverges by the alteratig series test. If f(x) 4 x, the the graph of f is positive, decreasig to 0, ad cocave up for x. For odd, iequality (7) implies that S < S < S +. (8) If we wated to estimate the value of S with error less tha 0.000, the typical method usig the error boud from iequality (5) would use a value of for which < This would require usig 0,000 terms. O the basis of iequality (8), however, we ca take as a estimate for S the midpoit of that iterval, that is, for odd, S S ( + + ) S 4 4, (9) with a error less tha half the width of the iterval. So for a error less tha 0.000, we oly eed ( ) + 4 < The first odd solutio is 7, just a bit less tha 0,000! The estimate from (9) usig 7 is S 3.459, with error less tha Sice S π, this estimate is actually withi of the true value. By the way, the partial sum S 7 is approximately

5 Example 7: S ( ) 8 ()! 0 This is a alteratig series that coverges by the alteratig series test. Let b +. It is ot obvious that the sequece b decreases mootoically to 0. A ivestigatio with the table feature of a graphig calculator, however, suggests that this is true for 3. We ca therefore use iequality (6) whe is a odd iteger greater tha 3 (ote that iequality (6) holds for odd s because this series starts with 0.) Hece S ( + )! < S < S + 8 ()! for odd 3. With 9 we ca estimate that S lies i the iterval ( , ), a iterval of legth But wait, we ca actually do better tha this! Sice the terms of this series decrease so quickly because of the factorial i the deomiator, we actually have + < for 3. So if we combie iequalities (5) ad (6), we ca deduce that for this series, 8 + S + ( + )! < S < S + 8+ ( + )! for odd 3. Now 9 gives the iterval ( , ) cotaiig the value of S, a iterval of legth (Note: What is the exact sum of this series?) Refereces [] Bart Brade, Calculatig Sums of Ifiite Series, The America Mathematical Mothly, Vol. 99, No. 7. (Aug. Sep., 99), [] Philip Calabrese, A Note o Alteratig Series, The America Mathematical Mothly, Vol. 69, No. 3. (Mar., 96), 5 7. (Reprited i Selected Papers o Calculus, The Mathematical Associatio of America, 968, ) [3] Rick Kremiski, Usig Simpso s Rule to Approximate Sums of Ifiite Series, The College Mathematics Joural, Vol. 8, No. 5. (Nov., 997), [4] R.K. Morley, The Remaider i Computig by Series, The America Mathematical Mothly, Vol. 57, No. 8. (Oct., 950), [5] R.K. Morley, Further Note o the Remaider i Computig by Series, The America Mathematical Mothly, Vol. 58, No. 6. (Ju. Jul., 95), Appedix Proof of Iequality () Let S ad let S a k. Suppose f() where the graph of f is positive, decreasig k to 0, ad cocave up, ad the improper itegral f(x) dx coverges. The series coverges by the itegral test. Because the graph is cocave up, the area of the shaded trapezoid of width show i Figure () is greater tha the area uder the curve. Therefore For egative edpoits, roud dow the left edpoit ad trucate the right edpoit. 5

6 Figure Figure Hece f(x) dx < (+ + + ). f(x) dx < (+ + + ) + ( ) + ( ) S S + I Figure (), the graph of f lies above that taget lie at x + (because of the positive cocavity) ad therefore also lies above the cotiuatio of the secat lie betwee x + ad x +. This implies that the area of the shaded trapezoid i Figure () of width betwee x ad x + is less tha the area uder the curve, ad so Hece + f(x) dx > + + (+ + ). f(x) dx > + + (+ + ) (+ +3 ) (+3 +4 ) S S Proof of Iequalities () ad (3) Let S ad let S k + a k. Suppose ( ) is a positive decreasig sequece ad lim L <, where the ratios decrease to L. The series coverges by the ratio test. 6

7 Let r + <. The a k+ a k We therefore coclude that S S k+ < r for all k. Hece a k + < r + < + r < r +3 < + r < r 3. +k < k r k r r + + k. But we also have L < a k+ a k for all k. By a similar argumet as above, S S k+ a k +k > k L k L L. Combiig these two results gives iequality (). A similar argumet for the iequalities with r ad L reversed proves iequality (3). Proof of Iequalities (6) ad (7) Let S ( ) + ad let S k k ( ) k+ a k, where ( ) is positive decreasig sequece that coverges to 0. Let b +, where we assume that the sequece (b ) also decreases mootoically to 0. The ad Because the sequece (b ) decreases, S S + ( ) (b + + b +3 + b +5 + ) S S + ( ) + (b + b + + b +4 + ). S S b + + b +3 + b +5 + < b + b + + b +4 + S S. Therefore S S < S S. Similarly, S S + < S S. But S lies betwee the successive partial sums, so it follows that ad Combiig these two results shows that S S S S + S S > S S + S + S S S + + S S < S S. + < S S < from which iequalities (6) ad (7) ca be obtaied. 7

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