Page 1 Article D’Alembert s Ratio Test Statement If ?u is a series of positive terms such that (a) lim u u then (i) ?u is convergent if (ii) ?u is divergent if (iii) ?u may converge or diverge if (i e ,the test fails if ) (b) lim u u ,then the series ?u is convergent Proof (a) ?u is series of positive terms u n u u lim u u Since lim u u ,therefore,for each a positive integer m such that | u u | n m u u Replacing n by m,m ,m , ,n in the above inequality,we have u u u u u u u u Multiplying the above (n m) inequality,we have ( ) u u ( ) ( ) (i) Let Choose such that (This is always possible since we have only to choose an such that From ( ),we have ( ) u u u u ( ) u u ( ) ( ) u K ( ) n m where K u ( ) Now,? ( ) being a geometric series with common ratio is convergent Therefore, Page 2 Article D’Alembert s Ratio Test Statement If ?u is a series of positive terms such that (a) lim u u then (i) ?u is convergent if (ii) ?u is divergent if (iii) ?u may converge or diverge if (i e ,the test fails if ) (b) lim u u ,then the series ?u is convergent Proof (a) ?u is series of positive terms u n u u lim u u Since lim u u ,therefore,for each a positive integer m such that | u u | n m u u Replacing n by m,m ,m , ,n in the above inequality,we have u u u u u u u u Multiplying the above (n m) inequality,we have ( ) u u ( ) ( ) (i) Let Choose such that (This is always possible since we have only to choose an such that From ( ),we have ( ) u u u u ( ) u u ( ) ( ) u K ( ) n m where K u ( ) Now,? ( ) being a geometric series with common ratio is convergent Therefore, For more notes, call 8130648819 by comparison test,the series ? u is convergent (ii) Let Choose such that (This is always possiblle since we have only to choose an such that Since , From ( ),we have u u ( ) u u ( ) u u ( ) ( ) u ( ) n m where u ( ) Now,? ( ) being a geometric series with common ratio is divergent Therefore, by comparison test,the series ? u is divergent (iii) Let First consider the series ? n n u n , u n u u n n n so that lim u u Since the series ? n is divergent,we find that if ,a series may diverge Next,consider the series ? n n u n , u (n ) u u (n ) n ( n n ) ( n ) so that Lim u u Since the series ? n is convergent,we find that if ,a series may converge The above two examples show that if ,a series may converge or diverge Hence the test fails when Remar Another equivalent form of Ratio Test is as follow If ?u is a positive term series such that lim u u then (i) ?u is convergent if (ii) ?u is divergent if Ex Discuss the convergence of the following series (i) (ii) ,(p ) (iii) (iv) Solution (i) Here u n n u (n ) (n ) Page 3 Article D’Alembert s Ratio Test Statement If ?u is a series of positive terms such that (a) lim u u then (i) ?u is convergent if (ii) ?u is divergent if (iii) ?u may converge or diverge if (i e ,the test fails if ) (b) lim u u ,then the series ?u is convergent Proof (a) ?u is series of positive terms u n u u lim u u Since lim u u ,therefore,for each a positive integer m such that | u u | n m u u Replacing n by m,m ,m , ,n in the above inequality,we have u u u u u u u u Multiplying the above (n m) inequality,we have ( ) u u ( ) ( ) (i) Let Choose such that (This is always possible since we have only to choose an such that From ( ),we have ( ) u u u u ( ) u u ( ) ( ) u K ( ) n m where K u ( ) Now,? ( ) being a geometric series with common ratio is convergent Therefore, For more notes, call 8130648819 by comparison test,the series ? u is convergent (ii) Let Choose such that (This is always possiblle since we have only to choose an such that Since , From ( ),we have u u ( ) u u ( ) u u ( ) ( ) u ( ) n m where u ( ) Now,? ( ) being a geometric series with common ratio is divergent Therefore, by comparison test,the series ? u is divergent (iii) Let First consider the series ? n n u n , u n u u n n n so that lim u u Since the series ? n is divergent,we find that if ,a series may diverge Next,consider the series ? n n u n , u (n ) u u (n ) n ( n n ) ( n ) so that Lim u u Since the series ? n is convergent,we find that if ,a series may converge The above two examples show that if ,a series may converge or diverge Hence the test fails when Remar Another equivalent form of Ratio Test is as follow If ?u is a positive term series such that lim u u then (i) ?u is convergent if (ii) ?u is divergent if Ex Discuss the convergence of the following series (i) (ii) ,(p ) (iii) (iv) Solution (i) Here u n n u (n ) (n ) For more notes, call 8130648819 u u n n (n ) (n ) n n (n ) (n )n (n ) n ( n n ) ( n ) lim u u lim ( n ) e , e - y D’Alembert’s Ratio Test, u is convergent (ii) Do yourself ,Ans Convergent- (iii) Do yourself ,Ans Convergent- (iv) Do yourself ,Ans Convergent- JAM MCQ Que Let S be the series ? ( ) ( ) and T be the series ? 4 5 ( ) of real numbers Then,which one of the following is true (a) oth the series S and T are convergent (b) S is convergent and T is divergent (c) S is divergent and T is convergent (d) oth the series S and T are divergent Sol S ? ( ) ( ) ? convergent or Consider the u for S u ( n ) ( ) u , (n ) - , ( ) - , n -, - u u ( n )( ) ( n )( ) n. n /( ) n. n /( ) . n / . n / lim u u by ratio test S is convergent Consider the u for T u ( n n ) ( ) u ( n n ) ( ) u u ( n n ) ( n n ) : n n ; : n n ; lim u u tends to So, T is divergent Ex Test the convergence of the following series (i) (ii) (iii) (iv) (v) Solution (i) Here u (n) u (n ) u u (n ) n n n n n n n Page 4 Article D’Alembert s Ratio Test Statement If ?u is a series of positive terms such that (a) lim u u then (i) ?u is convergent if (ii) ?u is divergent if (iii) ?u may converge or diverge if (i e ,the test fails if ) (b) lim u u ,then the series ?u is convergent Proof (a) ?u is series of positive terms u n u u lim u u Since lim u u ,therefore,for each a positive integer m such that | u u | n m u u Replacing n by m,m ,m , ,n in the above inequality,we have u u u u u u u u Multiplying the above (n m) inequality,we have ( ) u u ( ) ( ) (i) Let Choose such that (This is always possible since we have only to choose an such that From ( ),we have ( ) u u u u ( ) u u ( ) ( ) u K ( ) n m where K u ( ) Now,? ( ) being a geometric series with common ratio is convergent Therefore, For more notes, call 8130648819 by comparison test,the series ? u is convergent (ii) Let Choose such that (This is always possiblle since we have only to choose an such that Since , From ( ),we have u u ( ) u u ( ) u u ( ) ( ) u ( ) n m where u ( ) Now,? ( ) being a geometric series with common ratio is divergent Therefore, by comparison test,the series ? u is divergent (iii) Let First consider the series ? n n u n , u n u u n n n so that lim u u Since the series ? n is divergent,we find that if ,a series may diverge Next,consider the series ? n n u n , u (n ) u u (n ) n ( n n ) ( n ) so that Lim u u Since the series ? n is convergent,we find that if ,a series may converge The above two examples show that if ,a series may converge or diverge Hence the test fails when Remar Another equivalent form of Ratio Test is as follow If ?u is a positive term series such that lim u u then (i) ?u is convergent if (ii) ?u is divergent if Ex Discuss the convergence of the following series (i) (ii) ,(p ) (iii) (iv) Solution (i) Here u n n u (n ) (n ) For more notes, call 8130648819 u u n n (n ) (n ) n n (n ) (n )n (n ) n ( n n ) ( n ) lim u u lim ( n ) e , e - y D’Alembert’s Ratio Test, u is convergent (ii) Do yourself ,Ans Convergent- (iii) Do yourself ,Ans Convergent- (iv) Do yourself ,Ans Convergent- JAM MCQ Que Let S be the series ? ( ) ( ) and T be the series ? 4 5 ( ) of real numbers Then,which one of the following is true (a) oth the series S and T are convergent (b) S is convergent and T is divergent (c) S is divergent and T is convergent (d) oth the series S and T are divergent Sol S ? ( ) ( ) ? convergent or Consider the u for S u ( n ) ( ) u , (n ) - , ( ) - , n -, - u u ( n )( ) ( n )( ) n. n /( ) n. n /( ) . n / . n / lim u u by ratio test S is convergent Consider the u for T u ( n n ) ( ) u ( n n ) ( ) u u ( n n ) ( n n ) : n n ; : n n ; lim u u tends to So, T is divergent Ex Test the convergence of the following series (i) (ii) (iii) (iv) (v) Solution (i) Here u (n) u (n ) u u (n ) n n n n n n n For more notes, call 8130648819 lim u u lim n n n y D Alembert s Ratio Test,?u is divergent (ii) The given series is Here u n u (n ) u u n (n ) n n . n / . n / lim u u lim . n / y D Alembert s Ratio Test, u s convergent (iii) Do yourself ,Ans Divergent- (iv) u n , u (n ) divergent u u n (n ) n lim u u y D Alembert Ratio Test,the series is divergent (v) Do yourself ,Ans Divergent- Ex Examine the convergence or divergence of the following series (i) (ii) (iii) ? Solution (i) Here u n , (n ) - n ( n ) u n(n ) ( n )( n ) u u n ( n ) ( n )( n ) n(n ) u u ( n ) n n n lim u u lim n n y D Alembert’s Ratio Test, u is convergent (ii) Here u u u u . / . / lim u u lim y D Alembert’s Ratio Test, u is divergent (iii) Do yourself ,Ans Convergent- Ex Test the convergence of the following series (i) ? n (ii) ? n a a (iii) ? n (n ) n (iv) ? n n Solution (i) Do yourself ,Ans Convergent- Page 5 Article D’Alembert s Ratio Test Statement If ?u is a series of positive terms such that (a) lim u u then (i) ?u is convergent if (ii) ?u is divergent if (iii) ?u may converge or diverge if (i e ,the test fails if ) (b) lim u u ,then the series ?u is convergent Proof (a) ?u is series of positive terms u n u u lim u u Since lim u u ,therefore,for each a positive integer m such that | u u | n m u u Replacing n by m,m ,m , ,n in the above inequality,we have u u u u u u u u Multiplying the above (n m) inequality,we have ( ) u u ( ) ( ) (i) Let Choose such that (This is always possible since we have only to choose an such that From ( ),we have ( ) u u u u ( ) u u ( ) ( ) u K ( ) n m where K u ( ) Now,? ( ) being a geometric series with common ratio is convergent Therefore, For more notes, call 8130648819 by comparison test,the series ? u is convergent (ii) Let Choose such that (This is always possiblle since we have only to choose an such that Since , From ( ),we have u u ( ) u u ( ) u u ( ) ( ) u ( ) n m where u ( ) Now,? ( ) being a geometric series with common ratio is divergent Therefore, by comparison test,the series ? u is divergent (iii) Let First consider the series ? n n u n , u n u u n n n so that lim u u Since the series ? n is divergent,we find that if ,a series may diverge Next,consider the series ? n n u n , u (n ) u u (n ) n ( n n ) ( n ) so that Lim u u Since the series ? n is convergent,we find that if ,a series may converge The above two examples show that if ,a series may converge or diverge Hence the test fails when Remar Another equivalent form of Ratio Test is as follow If ?u is a positive term series such that lim u u then (i) ?u is convergent if (ii) ?u is divergent if Ex Discuss the convergence of the following series (i) (ii) ,(p ) (iii) (iv) Solution (i) Here u n n u (n ) (n ) For more notes, call 8130648819 u u n n (n ) (n ) n n (n ) (n )n (n ) n ( n n ) ( n ) lim u u lim ( n ) e , e - y D’Alembert’s Ratio Test, u is convergent (ii) Do yourself ,Ans Convergent- (iii) Do yourself ,Ans Convergent- (iv) Do yourself ,Ans Convergent- JAM MCQ Que Let S be the series ? ( ) ( ) and T be the series ? 4 5 ( ) of real numbers Then,which one of the following is true (a) oth the series S and T are convergent (b) S is convergent and T is divergent (c) S is divergent and T is convergent (d) oth the series S and T are divergent Sol S ? ( ) ( ) ? convergent or Consider the u for S u ( n ) ( ) u , (n ) - , ( ) - , n -, - u u ( n )( ) ( n )( ) n. n /( ) n. n /( ) . n / . n / lim u u by ratio test S is convergent Consider the u for T u ( n n ) ( ) u ( n n ) ( ) u u ( n n ) ( n n ) : n n ; : n n ; lim u u tends to So, T is divergent Ex Test the convergence of the following series (i) (ii) (iii) (iv) (v) Solution (i) Here u (n) u (n ) u u (n ) n n n n n n n For more notes, call 8130648819 lim u u lim n n n y D Alembert s Ratio Test,?u is divergent (ii) The given series is Here u n u (n ) u u n (n ) n n . n / . n / lim u u lim . n / y D Alembert s Ratio Test, u s convergent (iii) Do yourself ,Ans Divergent- (iv) u n , u (n ) divergent u u n (n ) n lim u u y D Alembert Ratio Test,the series is divergent (v) Do yourself ,Ans Divergent- Ex Examine the convergence or divergence of the following series (i) (ii) (iii) ? Solution (i) Here u n , (n ) - n ( n ) u n(n ) ( n )( n ) u u n ( n ) ( n )( n ) n(n ) u u ( n ) n n n lim u u lim n n y D Alembert’s Ratio Test, u is convergent (ii) Here u u u u . / . / lim u u lim y D Alembert’s Ratio Test, u is divergent (iii) Do yourself ,Ans Convergent- Ex Test the convergence of the following series (i) ? n (ii) ? n a a (iii) ? n (n ) n (iv) ? n n Solution (i) Do yourself ,Ans Convergent- For more notes, call 8130648819 (ii) Here u n a a u (n ) a a u u (n a)( a) ((n ) a)( a) n . a n / n 6. n / a n 7 . a / . a / a n . n / a n . a / a lim u u y D Alembert s Ratio Test, u is convergent (iii) Do yourself ,Ans Convergent- (iv) Do yourself ,Ans Convergent- Ex Discuss the convergence or divergence of the following series (i) ? n n (ii) ? n n (iii) ? n n Solution (i) Here u n n u (n ) (n ) (n ) (n )n n n u u n n n n lim u u lim n n y D Alembert s Ratio Test, u is convergent (ii) Do yourself ,Ans Convergent- (iii) Do yourself ,Ans Divergent- Ex Discuss the convergence or divergence of the following series (i) ? x n ,x (ii) ? x a vn (iii) ? v n n x (iv) ? vn vn x (v) ? x n ,x (vi) ? n n x ,x Solution (i) Here u x n u x (n ) u u x x (n ) n x ( n ) lim u u lim x ( n ) x y D Alembert s Ratio Test, u converges if x i e x and diverges if x i e ,x When x , lim u u Ratio Test Fails Now,when x , u n n ?u ? n is of the form ? n with p ?u is convergent Hence the given series u converges if x and diverges if x (ii) Do yourself ,Ans converges if x and diverges if x -Read More

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