Tính giới hạn \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaacM % gacaGGTbWaamWaaeaadaWcaaqaaiaaigdaaeaacaaIXaGaaiOlaiaa % isdaaaGaey4kaSYaaSaaaeaacaaIXaaabaGaaGOmaiaac6cacaaI1a % aaaiabgUcaRiaac6cacaGGUaGaaiOlaiabgUcaRmaalaaabaGaaGym % aaqaaiaad6gacaGGOaGaamOBaiabgUcaRiaaiodacaGGPaaaaaGaay % 5waiaaw2faaaaa!4B08! \lim \left[ {\frac{1}{{1.4}} + \frac{1}{{2.5}} + ... + \frac{1}{{n(n + 3)}}} \right]\)
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Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaacM % gacaGGTbWaamWaaeaadaWcaaqaaiaaigdaaeaacaaIXaGaaiOlaiaa % isdaaaGaey4kaSYaaSaaaeaacaaIXaaabaGaaGOmaiaac6cacaaI1a % aaaiabgUcaRiaac6cacaGGUaGaaiOlaiabgUcaRmaalaaabaGaaGym % aaqaaiaad6gacaGGOaGaamOBaiabgUcaRiaaiodacaGGPaaaaaGaay % 5waiaaw2faaiabg2da9iGacYgacaGGPbGaaiyBamaadmaabaWaaSaa % aeaacaaIXaaabaGaaG4maaaadaqadaqaaiaaigdacqGHsisldaWcaa % qaaiaaigdaaeaacaaI0aaaaiabgUcaRmaalaaabaGaaGymaaqaaiaa % ikdaaaGaeyOeI0YaaSaaaeaacaaIXaaabaGaaGynaaaacqGHRaWkda % WcaaqaaiaaigdaaeaacaaIZaaaaiabgkHiTmaalaaabaGaaGymaaqa % aiaaiAdaaaGaey4kaSIaaiOlaiaac6cacaGGUaGaey4kaSYaaSaaae % aacaaIXaaabaGaamOBaaaacqGHsisldaWcaaqaaiaaigdaaeaacaWG % UbGaey4kaSIaaG4maaaaaiaawIcacaGLPaaaaiaawUfacaGLDbaaaa % a!6AB6! \lim \left[ {\frac{1}{{1.4}} + \frac{1}{{2.5}} + ... + \frac{1}{{n(n + 3)}}} \right] = \lim \left[ {\frac{1}{3}\left( {1 - \frac{1}{4} + \frac{1}{2} - \frac{1}{5} + \frac{1}{3} - \frac{1}{6} + ... + \frac{1}{n} - \frac{1}{{n + 3}}} \right)} \right]\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0Jaci % iBaiaacMgacaGGTbWaamWaaeaadaWcaaqaaiaaigdaaeaacaaIZaaa % amaabmaabaGaaGymaiabgUcaRmaalaaabaGaaGymaaqaaiaaikdaaa % Gaey4kaSYaaSaaaeaacaaIXaaabaGaaG4maaaacqGHsisldaWcaaqa % aiaaigdaaeaacaWGUbGaey4kaSIaaGymaaaacqGHsisldaWcaaqaai % aaigdaaeaacaWGUbGaey4kaSIaaGOmaaaacqGHsisldaWcaaqaaiaa % igdaaeaacaWGUbGaey4kaSIaaG4maaaaaiaawIcacaGLPaaaaiaawU % facaGLDbaaaaa!5135! = \lim \left[ {\frac{1}{3}\left( {1 + \frac{1}{2} + \frac{1}{3} - \frac{1}{{n + 1}} - \frac{1}{{n + 2}} - \frac{1}{{n + 3}}} \right)} \right]\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaS % aaaeaacaaIXaGaaGymaaqaaiaaigdacaaI4aaaaiabgkHiTiGacYga % caGGPbGaaiyBamaadmaabaWaaSaaaeaacaaIZaGaamOBamaaCaaale % qabaGaaGOmaaaakiabgUcaRiaaigdacaaIYaGaamOBaiabgUcaRiaa % igdacaaIXaaabaWaaeWaaeaacaWGUbGaey4kaSIaaGymaaGaayjkai % aawMcaamaabmaabaGaamOBaiabgUcaRiaaikdaaiaawIcacaGLPaaa % daqadaqaaiaad6gacqGHRaWkcaaIZaaacaGLOaGaayzkaaaaaaGaay % 5waiaaw2faaiabg2da9maalaaabaGaaGymaiaaigdaaeaacaaIXaGa % aGioaaaaaaa!5859! = \frac{{11}}{{18}} - \lim \left[ {\frac{{3{n^2} + 12n + 11}}{{\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}} \right] = \frac{{11}}{{18}}\)