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Convergence of series
Divergence, ratio, comparison and integral tests, and the harmonic series.
An infinite series converges when its partial sums approach a limit. The terms going to zero is necessary but not enough. The harmonic series 1 + ½ + ⅓ + … diverges even though its terms vanish. The ratio test settles most series with factorials or powers; comparison and the integral test handle the p-series family.
Çözümlü örnek · does 1/n^2 for n = 1 to oo converge
Adım adım
- \sum_{n=1}^{\infty} \frac{1}{n^{2}}
Does this series converge?
- \lim_{n\to\infty} \frac{1}{n^{2}} = 0
Divergence test: if the terms do not go to 0 the series cannot converge.
- \lim \left|\frac{a_{k+1}}{a_k}\right| = \lim \left|{\frac{n^{2}}{\left(n + 1\right)^{2}}}\right| = 1
Ratio test: < 1 converges, > 1 diverges, = 1 says nothing.
- = \frac{\pi^{2}}{6}
It converges, and the CAS knows the sum.
Cevabı göster
Şimdi sen Bir problem seçin, ya da kendi probleminizi yazın ya da çizin. Her adım, bir resim, cevap sorulana kadar gizli.
Burada kullanılan sembolleri
Tüm tanım, resim ve her harfin anlamı için herhangi bir sembolü tıklayın.
Nasıl: Convergence of series
- Does this series converge?
- Divergence test: if the terms do not go to 0 the series cannot converge.
- Ratio test: < 1 converges, > 1 diverges, = 1 says nothing.
- It converges, and the CAS knows the sum.
Sık sorulan sorular
What is the ε, δ definition actually saying?
That you can make the output as close to the limit as anyone demands (within ε) by keeping the input close enough (within δ). It replaces "approaches" with a challenge-and-response that can be checked.
Why does the harmonic series diverge when its terms go to zero?
Because the terms shrink too slowly: group them as 1/3 + 1/4 > 1/2, 1/5 + … + 1/8 > 1/2, and so on: infinitely many halves.
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