Title: Suppenemistestit

Do the following series converge? Suppenevatko seuraavat sarjat?

(a) \[\sum_{n=1}^{\infty} \frac{n^2}{[value]^n} \] (b) \[\sum_{n=1}^{\infty} \frac{(n+1)[value]^n}{n!} \] (c) \[\sum_{n=1}^{\infty} \frac{n}{n^[value]+[value]n+[value]} \]

"1" = Converges.Suppenee.

"0" = Does not converge.Ei suppene.

Write the number corresponding to your answer on the boxes below.Kirjoita oikeaa ratkaisua vastaava numero alla oleviin tekstikenttiin.

(a)

(b)

(c)
Feedback: Model solution: Mallivastaus: (a) \[\sum_{n=1}^{\infty} \frac{n^2}{[value]^n} \] Using the ratio test: Suhdetestillä: \begin{eqnarray*} \frac{a_{n+1}}{a_n} & = & \frac{ \frac{(n+1)^2}{[value]^{n+1}}}{ \frac{n^2}{[value]^n}} \\ & = & \frac{1}{[value]} \frac{n^2+2n+1}{n^2} \\ & = & \frac{1}{[value]} \frac{1+\frac{2}{n}+\frac{1}{n^2}}{1} \\ \\ & \to & \frac{1}{[value]}, \ \text{whenkun} \ n\to \infty. \end{eqnarray*} Because \(\lim_{n\to \infty} \frac{a_{n+1}}{a_n}=\frac{1}{[value]} < 1\), the series converges.Koska \(\lim_{n\to \infty} \frac{a_{n+1}}{a_n}=\frac{1}{[value]} < 1\), sarja suppenee. (b) \[\sum_{n=1}^{\infty} \frac{(n+1)[value]^n}{n!} \] Using the ratio test: Suhdetestillä: \begin{eqnarray*} \frac{a_{n+1}}{a_n} & = & \frac{(n+2)[value]^{n+1}}{(n+1)!}\cdot \frac{n!}{(n+1)[value]^n} \\ & = & [value]\cdot \frac{(n+2)n!}{(n+1)n!(n+1)} \\ & = & [value]\cdot \frac{n+2}{n^2+2n+1} \\ & = & [value]\cdot \frac{\frac{1}{n}+\frac{2}{n^2}}{1+\frac{2}{n}+\frac{1}{n^2}} \\ & \to & [value]\cdot \frac{0}{1}=0, \ \text{whenkun} \ n\to \infty, \end{eqnarray*} so the series converges.joten sarja suppenee. (c) \[\sum_{n=1}^{\infty} \frac{n}{n^{[value]}+[value]n+[value]} \] \[a_n=\frac{n}{n^{[value]}+[value]n+[value]}\leq \frac{n}{n^{[value]}}=\frac{1}{n^[value]}. \] Being over-harmonic, the series \(\sum_{n=1}^{\infty} \frac{1}{n^{[value]}}\) converges, so from the comparison test we can conclude that \(\sum_{n=1}^{\infty} \frac{n}{n^{[value]}+[value]n+[value]}\) converges.Sarja \(\sum_{n=1}^{\infty} \frac{1}{n^{[value]}}\) suppenee yliharmonisena sarjana, joten vertailuperiaatteen nojalla sarja \(\sum_{n=1}^{\infty} \frac{n}{n^{[value]}+[value]n+[value]}\) suppenee.

Applying root or ratio test, examine whether the following series converges: Tutki suhde- tai juuritestin avulla, suppeneeko seuraava sarja:

\[\sum_{n=1}^\infty [value] \]

The limit of the test isTestin raja-arvo on |  |

Does the series converge or diverge? Type the number corresponding to your answer: Hajaantuuko vai suppeneeko sarja? Valitse seuraavista vaihtoehdoista:

(1) | DivergesHajaantuu
(2) | ConvergesSuppenee
Feedback: Model solution: Mallivastaus: Method ITapa I According to ratio test, the series \(\sum_{n=1}^\infty a_n\) converges, ifSuhdetestin mukaan sarja \(\sum_{n=1}^\infty a_n\) suppenee, jos pätee \[ \lim_{n\to\infty}\frac{a_{n+1}}{a_n} < 1. \] \[[value]\cdot\frac{1}{[value]} = [value] \to [value] \text{, whenkun } n\to\infty\] Method IITapa II According to root test, the series converges, ifJuuritestin mukaan sama sarja suppenee, jos \[\lim_{n\to\infty}\sqrt[n]{a_n} <1. \] \[\sqrt[n][value] = \frac{[value]^{ [value] }}{[value]^{ [value] }}\sqrt[n][value] \to [value] \text{, whenkun } n\to\infty \]

Tutkitaan sarjan \[\sum\limits^\infty_{k=1} [value]\] suppenemista/hajaantumista integraalitestin avulla.

Funktio \[f(x)=[value]\]on jatkuva, positiivinen ja vähenevä, kun \(x\ge1\).

Laske epäoleellisen integraalin tarkka arvo \[\int\limits_1^{\infty}f(x)\,\mathrm{d}x\] ja päättele, suppeneeko vai hajaantuuko tutkittava sarja. \(\infty\) merkitään inf.

Consider the convergence or divergence of the series \[\sum\limits^\infty_{k=1} [value]\] with the help of the integral test.

The function \[f(x)=[value]\] is continuous, positive and decreasing when \(x\ge1\).

Find the exact value of the improper integral \[\int\limits_1^{\infty}f(x)\,\mathrm{d}x\] and conclude, whether the series converges or diverges. \(\infty\) is written as inf.

Epäoleellisen integraalin tarkka arvo onThe exact value of the improper integral is .

Väite: Sarja suppenee.Claim: The series converges.
Feedback: Malliratkaisu: Epäoleellinen integraali \[\int\limits_{1}^{\infty} f(x)\, \mathrm{d}x=\int\limits_{1}^{\infty}[value] \, \mathrm{d}x ={\Bigl/}_{\!\!1}^{\,\infty}(-[value]) =[value]<\infty.\] Sarja suppenee integraalitestin nojalla. Model solution: The improper integral \[\int\limits_{1}^{\infty} f(x)\, \mathrm{d}x=\int\limits_{1}^{\infty}[value] \, \mathrm{d}x ={\Bigl/}_{\!\!1}^{\,\infty}(-[value]) =[value]<\infty.\] The series converges by the integral test.

Tutkitaan sarjan \[\sum\limits^\infty_{k=1} [value]\] suppenemista/hajaantumista integraalitestin avulla.

Funktio \[f(x)=[value]\]on jatkuva, positiivinen ja vähenevä, kun \(x\ge1\).

Laske epäoleellisen integraalin tarkka arvo \[\int\limits_1^{\infty}f(x)\,\mathrm{d}x\] ja päättele, suppeneeko vai hajaantuuko tutkittava sarja. \(\infty\) merkitään inf.

Consider the convergence or divergence of the series \[\sum\limits^\infty_{k=1} [value]\] with the help of the integral test.

The function \[f(x)=[value]\] is continuous, positive and decreasing when \(x\ge1\).

Find the exact value of the improper integral \[\int\limits_1^{\infty}f(x)\,\mathrm{d}x\] and conclude, whether the series converges or diverges. \(\infty\) is written as inf.

Epäoleellisen integraalin tarkka arvo onThe exact value of the improper integral is .

Väite: Sarja suppenee.Claim: The series converges.
Feedback: Malliratkaisu: Epäoleellinen integraali \[\int\limits_{1}^{\infty} f(x)\, dx=\int\limits_{1}^{\infty}[value] \, \mathrm{d}x ={\Bigl/}_{\!\!1}^{\,\infty}(-[value]) =[value]<\infty.\] Sarja suppenee integraalitestin nojalla. Model solution: The improper integral \[\int\limits_{1}^{\infty} f(x)\, dx=\int\limits_{1}^{\infty}[value] \, \mathrm{d}x ={\Bigl/}_{\!\!1}^{\,\infty}(-[value]) =[value]<\infty.\] The series converges by the integral test.

Tutkitaan sarjan \[\sum\limits^\infty_{k=1}\frac{\ln(k+[value])}{k+[value]}\] suppenemista/hajaantumista integraalitestin avulla.

Funktio \[f(x)=\frac{\ln(x+[value])}{x+[value]}\] on jatkuva ja positiivinen, kun \(x\ge1\). Funktion vähenevyys voidaan osoittaa derivaattafunktion \(f'(x)\) avulla. Luonnollinen logaritmi merkitään ln, esim. ln(2).

Consider the convergence or divergence of the series \[\sum\limits^\infty_{k=1} \frac{\ln(k+[value])}{k+[value]}\] with the help of the integral test.

The function \[f(x)=\frac{\ln(x+[value])}{x+[value]}\] is continuous and positive when \(x\ge1\). The decreasing property of the function can be shown by the derivative function \(f'(x)\). The natural logarithm is written as ln, e.g. ln(3).

Derivaattafunktio \(f'(x)\) onThe derivative function \(f'(x)\) is .

Laske epäoleellisen integraalin tarkka arvo \[\int\limits_1^{\infty}f(x)\,\mathrm{d}x\] ja päättele, suppeneeko vai hajaantuuko tutkittava sarja. \(\infty\) merkitään inf.

Find the exact value of the improper integral \[\int\limits_1^{\infty}f(x)\,\mathrm{d}x\] and conclude, whether the series converges or diverges. \(\infty\) is written as inf.

Epäoleellisen integraalin tarkka arvo on The exact value of the improper integral is .

Väite: Sarja hajaantuu. Claim: The series diverges.
Feedback: Malliratkaisu: \[ f'(x)=\frac{\frac{1}{x+[value]}\cdot(x+[value])- \ln(x+[value])\cdot1}{(x+[value])^2}=\frac{1- \ln(x+[value])}{(x+[value])^2} \] Epäoleellinen integraali \[\int\limits_{1}^{\infty} f(x)\, \mathrm{d}x=\int\limits_{1}^{\infty} \frac{\ln(x+[value])}{x+[value]} \, \mathrm{d}x ={\Bigl/}_{\!\!1}^{\,\infty}\frac12\ln^2\big|x+[value] \big| =\infty.\] Sarja hajaantuu integraalitestin nojalla. Model solution: \[ f'(x)=\frac{\frac{1}{x+[value]}\cdot(x+[value])- \ln(x+[value])\cdot1}{(x+[value])^2}=\frac{1- \ln(x+[value])}{(x+[value])^2} \] The improper integral \[\int\limits_{1}^{\infty} f(x)\, \mathrm{d}x=\int\limits_{1}^{\infty} \frac{\ln(x+[value])}{x+[value]} \, \mathrm{d}x ={\Bigl/}_{\!\!1}^{\,\infty}\frac12\ln^2\big|x+[value] \big| =\infty.\] The series diverges by the integral test.

Using ratio test, find outSelvitä suhdetestillä,

a) whether the series \(\sum_{n=0}^{\infty} \frac{[value]^n}{(n+1)\cdot[value]^n}\) convergesa) suppeneeko sarja \(\sum_{n=0}^{\infty} \frac{[value]^n}{(n+1)\cdot[value]^n}\).

The limit of the ratio test: Suhdetestin raja-arvo:

Interpretation of the ratio test: Suhdetestin tulkinta:

b) for which \(x\) does the series \(\sum_{n=0}^{\infty} \frac{n+1}{[value]^{n}} x^n\) converge? b) millä muuttujan \(x\) arvoilla sarja \(\sum_{n=0}^{\infty} \frac{n+1}{[value]^{n}} x^n\) suppenee.

\(x\in~\),
Feedback: Model solution: Mallivastaus: a) The convergence of the series \(\sum_{n=0}^{\infty} \frac{[value]^n}{(n+1)\cdot[value]^n}\)a) Sarjan \(\sum_{n=0}^{\infty} \frac{[value]^n}{(n+1)\cdot[value]^n}\) suppeneminen Applying the ratio test: Suhdetestillä: \[\left|\frac{\frac{[value]^{n+1}}{((n+1)+1)\cdot[value]^{n+1}}}{\frac{[value]^n}{(n+1)\cdot[value]^n}}\right| = \frac{[value](n+1)}{[value](n+2)} = \frac{[value](1+\frac{1}{n})}{[value](1+\frac{2}{n})} \to \frac{[value]}{[value]}\text{, whenkun }n\to\infty \] Since \(\frac{[value]}{[value]}<1\), the series converges.Koska \(\frac{[value]}{[value]}<1\), sarja suppenee. b) For which \(x\) does the series \(\sum_{n=0}^{\infty} \frac{n+1}{[value]^{n}} x^n\) converge? b) Millä muuttujan \(x\) arvoilla \(\sum_{n=0}^{\infty} \frac{n+1}{[value]^{n}} x^n\) suppenee? Applying the ratio test: Suhdetestillä: \[\left|\frac{\frac{(n+1)+1}{[value]^{n+1}} x^{n+1}}{\frac{n+1}{[value]^{n}} x^n}\right| =\left|\frac{n+2}{[value](n+1)}x\right| = \left|\frac{1+\frac{2}{n}}{[value](1+\frac{1}{n})}x\right| \to \frac{\left|x\right|}{[value]}\text{, whenkun } n\to\infty \] ThusTällöin \[\frac{\left|x\right|}{[value]}\leq 1 \Rightarrow \left|x\right| \leq [value]\] If \(x=\pm [value]\), we see that the series is of the form \(\pm\sum_{n=0}^{\infty}n+1\) and so does not converge. Therefore, the series converges if \(x\in ]-[value],[value][ \).Jos \(x=\pm [value]\), nähdään että sarja on muotoa \(\pm\sum_{n=0}^{\infty}n+1\), jolloin se ei suppene. Siis sarja suppenee kun \(x\in ]-[value],[value][ \).

Tutkitaan sarjan \[\sum\limits^\infty_{k=1}([value])\] suppenemista/hajaantumista osasummien jonon \[(S_n)^\infty_{n=1}, S_n=\sum\limits^n_{k=1}([value]), n=1, 2, 3, \ldots,\]avulla.

Määrää aluksi osasummien jonon yleisen termin \(S_n\) lauseke.

Consider the convergence or divergence of the series \[ \sum\limits^\infty_{k=1}([value]) \] with the help of the sequence of partial sums \[ (S_n)^\infty_{n=1}, S_n=\sum\limits^n_{k=1}([value]), n=1, 2, 3, \ldots. \]

At first find the expression for the general term \(S_n\) of the sequence of partial sums.

\(S_n=\) , \(n=1\), \(2\), \(3\), \(\ldots\)

Laske \(\lim\limits_{n\to\infty}S_n\) ja päättele, suppeneeko vai hajaantuuko tutkittava sarja.

Calculate \(\lim\limits_{n\to\infty}S_n\) and conclude, whether the series converges or diverges.

Väite: Sarja suppenee. Claim: The series converges.
Feedback: Malliratkaisu: Muodostetaan osasummien jonon yleisen termin lauseke: Model solution: Form the expression for the general term of the sequence of partial sums : \begin{align*} S_n &=\sum\limits_{k=1}^{n}([value]-[value])\\ &=([value]-[value])+([value]-[value])+([value]-[value])+ \cdots+ ([value]-[value])+([value]-[value])\\ &=[value]-[value],\, n=1,2,\ldots. \end{align*} Tällöin \( \lim\limits_{n\to\infty}S_n=\lim\limits_{n\to\infty}([value]-[value]) =-\infty\). Sarja siis hajaantuu. Hence \( \lim\limits_{n\to\infty}S_n=\lim\limits_{n\to\infty}([value]-[value]) =-\infty\). Therefore the series diverges.

Tutkitaan sarjan \[\sum\limits^\infty_{k=1}([value])\] suppenemista/hajaantumista osasummien jonon \[(S_n)^\infty_{n=1}, S_n=\sum\limits^n_{k=1}([value]), n=1, 2, 3, \ldots,\]avulla.

Määrää aluksi osasummien jonon yleisen termin \(S_n\) lauseke.

Consider the convergence or divergence of the series \[ \sum\limits^\infty_{k=1}([value]) \] with the help of the sequence of partial sums \[ (S_n)^\infty_{n=1}, S_n=\sum\limits^n_{k=1}([value]), n=1, 2, 3, \ldots. \]

At first find the expression for the general term \(S_n\) of the sequence of partial sums.

\(S_n=\) , \(n=1\), \(2\), \(3\), \(\ldots\)

Laske \(\lim\limits_{n\to\infty}S_n\) ja päättele, suppeneeko vai hajaantuuko tutkittava sarja.

Calculate \(\lim\limits_{n\to\infty}S_n\) and conclude, whether the series converges or diverges.

Väite: Sarja suppenee. Claim: The series converges.
Feedback: Malliratkaisu: Muodostetaan osasummien jonon yleisen termin lauseke: Model solution: Form the expression for the general term of the sequence of partial sums : \begin{align*} S_n &=\sum\limits_{k=1}^{n}([value]-[value])\\ &=([value]-[value])+([value]-[value])+([value]-[value]) + \cdots+ ([value]-[value])+([value]-[value])\\ &=[value]-[value],\, n=1,2,\ldots. \end{align*} Tällöin \( \lim\limits_{n\to\infty}S_n=\lim\limits_{n\to\infty}([value]-[value]) =[value]\). Sarja siis suppenee. Hence \( \lim\limits_{n\to\infty}S_n=\lim\limits_{n\to\infty}([value]-[value]) =[value]\). Therefore the series converges.

Tutkitaan sarjan \[\sum\limits^\infty_{k=1}([value])\] suppenemista/hajaantumista osasummien jonon \[(S_n)^\infty_{n=1}, S_n=\sum\limits^n_{k=1}([value]), n=1, 2, 3, \ldots,\]avulla.

Määrää aluksi osasummien jonon yleisen termin \(S_n\) lauseke.

Consider the convergence or divergence of the series \[ \sum\limits^\infty_{k=1}([value]) \] with the help of the sequence of partial sums \[ (S_n)^\infty_{n=1}, S_n=\sum\limits^n_{k=1}([value]), n=1, 2, 3, \ldots. \]

At first find the expression for the general term \(S_n\) of the sequence of partial sums.

\(S_n=\) , \(n=1\), \(2\), \(3\), \(\ldots\)

Laske \(\lim\limits_{n\to\infty}S_n\) ja päättele, suppeneeko vai hajaantuuko tutkittava sarja.

Calculate \(\lim\limits_{n\to\infty}S_n\) and conclude, whether the series converges or diverges.

Väite: Sarja suppenee. Claim: The series converges.
Feedback: Malliratkaisu: Muodostetaan osasummien jonon yleisen termin lauseke: Model solution: Form the expression for the general term of the sequence of partial sums : \begin{align*} S_n &=\sum\limits_{k=1}^{n}([value])\\ &=([value]^[value]-[value]^[value])+([value]^[value]-[value]^[value])+([value]^[value]-[value]^[value])+([value]^[value]-[value]^[value])+ \cdots+([value]-[value])+ ([value]-[value])+([value]-[value])\\ &=-[value]^[value]+[value] = -[value]+[value],\, n=1,2,\ldots. \end{align*} Tällöin \( \lim\limits_{n\to\infty}S_n=\lim\limits_{n\to\infty}(-[value]+[value]) =\infty\). Sarja siis hajaantuu. Hence \( \lim\limits_{n\to\infty}S_n=\lim\limits_{n\to\infty}(-[value]+[value]) =\infty\). Therefore the series diverges.